Show that the following sequence converges. Please Critique my proof.












6












$begingroup$


The problem is as follows:




Let ${a_n}$ be a sequence of nonnegative numbers such that
$$
a_{n+1}leq a_n+frac{(-1)^n}{n}.
$$
Show that $a_n$ converges.




My (wrong) proof:



Notice that
$$
|a_{n+1}-a_n|leq left|frac{(-1)^n}{n}right|leqfrac{1}{n}
$$
and since it is known that $frac{1}{n}rightarrow 0$ as $nrightarrow infty$. We see that we can arbitarily bound, $|a_{n+1}-a_n|$. Thus, $a_n$ converges.



My question:
This is a question from a comprehensive exam I found and am using to review.



Should I argue that we should select $N$ so that $n>N$ implies $left|frac{1}{n}right|<epsilon$ as well?



Notes: Currently working on the proof.






real-analysis sequences-and-series convergence fake-proofs







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Your proof is not correct. Your arguments would also work for $a_n = sum_{i=1}^n frac 1 i$, which does not converge.
    $endgroup$
    – Falrach
    2 hours ago






  • 1




    $begingroup$
    Note that you not only need to bound $left| a_{n+1} - a_n right|$ arbitrarily small, but also $left| a_{m} - a_n right|$ for all $m,n geq N$ (where $N$ can be chosen according to the bound).
    $endgroup$
    – Maximilian Janisch
    36 mins ago


















6












$begingroup$


The problem is as follows:




Let ${a_n}$ be a sequence of nonnegative numbers such that
$$
a_{n+1}leq a_n+frac{(-1)^n}{n}.
$$
Show that $a_n$ converges.




My (wrong) proof:



Notice that
$$
|a_{n+1}-a_n|leq left|frac{(-1)^n}{n}right|leqfrac{1}{n}
$$
and since it is known that $frac{1}{n}rightarrow 0$ as $nrightarrow infty$. We see that we can arbitarily bound, $|a_{n+1}-a_n|$. Thus, $a_n$ converges.



My question:
This is a question from a comprehensive exam I found and am using to review.



Should I argue that we should select $N$ so that $n>N$ implies $left|frac{1}{n}right|<epsilon$ as well?



Notes: Currently working on the proof.






real-analysis sequences-and-series convergence fake-proofs







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Your proof is not correct. Your arguments would also work for $a_n = sum_{i=1}^n frac 1 i$, which does not converge.
    $endgroup$
    – Falrach
    2 hours ago






  • 1




    $begingroup$
    Note that you not only need to bound $left| a_{n+1} - a_n right|$ arbitrarily small, but also $left| a_{m} - a_n right|$ for all $m,n geq N$ (where $N$ can be chosen according to the bound).
    $endgroup$
    – Maximilian Janisch
    36 mins ago
















6












6








6


1



$begingroup$


The problem is as follows:




Let ${a_n}$ be a sequence of nonnegative numbers such that
$$
a_{n+1}leq a_n+frac{(-1)^n}{n}.
$$
Show that $a_n$ converges.




My (wrong) proof:



Notice that
$$
|a_{n+1}-a_n|leq left|frac{(-1)^n}{n}right|leqfrac{1}{n}
$$
and since it is known that $frac{1}{n}rightarrow 0$ as $nrightarrow infty$. We see that we can arbitarily bound, $|a_{n+1}-a_n|$. Thus, $a_n$ converges.



My question:
This is a question from a comprehensive exam I found and am using to review.



Should I argue that we should select $N$ so that $n>N$ implies $left|frac{1}{n}right|<epsilon$ as well?



Notes: Currently working on the proof.






real-analysis sequences-and-series convergence fake-proofs







share|cite|improve this question











$endgroup$




The problem is as follows:




Let ${a_n}$ be a sequence of nonnegative numbers such that
$$
a_{n+1}leq a_n+frac{(-1)^n}{n}.
$$
Show that $a_n$ converges.




My (wrong) proof:



Notice that
$$
|a_{n+1}-a_n|leq left|frac{(-1)^n}{n}right|leqfrac{1}{n}
$$
and since it is known that $frac{1}{n}rightarrow 0$ as $nrightarrow infty$. We see that we can arbitarily bound, $|a_{n+1}-a_n|$. Thus, $a_n$ converges.



My question:
This is a question from a comprehensive exam I found and am using to review.



Should I argue that we should select $N$ so that $n>N$ implies $left|frac{1}{n}right|<epsilon$ as well?



Notes: Currently working on the proof.






real-analysis sequences-and-series convergence fake-proofs




real-analysis sequences-and-series convergence fake-proofs






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 34 mins ago









GNUSupporter 8964民主女神 地下教會

13.9k72650




13.9k72650










asked 2 hours ago









DarelDarel

1049




1049








  • 4




    $begingroup$
    Your proof is not correct. Your arguments would also work for $a_n = sum_{i=1}^n frac 1 i$, which does not converge.
    $endgroup$
    – Falrach
    2 hours ago






  • 1




    $begingroup$
    Note that you not only need to bound $left| a_{n+1} - a_n right|$ arbitrarily small, but also $left| a_{m} - a_n right|$ for all $m,n geq N$ (where $N$ can be chosen according to the bound).
    $endgroup$
    – Maximilian Janisch
    36 mins ago
















  • 4




    $begingroup$
    Your proof is not correct. Your arguments would also work for $a_n = sum_{i=1}^n frac 1 i$, which does not converge.
    $endgroup$
    – Falrach
    2 hours ago






  • 1




    $begingroup$
    Note that you not only need to bound $left| a_{n+1} - a_n right|$ arbitrarily small, but also $left| a_{m} - a_n right|$ for all $m,n geq N$ (where $N$ can be chosen according to the bound).
    $endgroup$
    – Maximilian Janisch
    36 mins ago










4




4




$begingroup$
Your proof is not correct. Your arguments would also work for $a_n = sum_{i=1}^n frac 1 i$, which does not converge.
$endgroup$
– Falrach
2 hours ago




$begingroup$
Your proof is not correct. Your arguments would also work for $a_n = sum_{i=1}^n frac 1 i$, which does not converge.
$endgroup$
– Falrach
2 hours ago




1




1




$begingroup$
Note that you not only need to bound $left| a_{n+1} - a_n right|$ arbitrarily small, but also $left| a_{m} - a_n right|$ for all $m,n geq N$ (where $N$ can be chosen according to the bound).
$endgroup$
– Maximilian Janisch
36 mins ago






$begingroup$
Note that you not only need to bound $left| a_{n+1} - a_n right|$ arbitrarily small, but also $left| a_{m} - a_n right|$ for all $m,n geq N$ (where $N$ can be chosen according to the bound).
$endgroup$
– Maximilian Janisch
36 mins ago












3 Answers
3






active

oldest

votes


















3












$begingroup$

Consider $b_n = a_n + sum_{k=1}^{n-1} frac{(-1)^{k-1}}{k}$. Then



$$ b_{n+1}
= a_{n+1} + sum_{k=1}^{n} frac{(-1)^{k-1}}{k}
leq a_n + frac{(-1)^n}{n} + sum_{k=1}^{n} frac{(-1)^{k-1}}{k}
= b_n, $$



which shows that $(b_n)$ is non-increasing. Moreover, since $sum_{k=1}^{infty} frac{(-1)^{k-1}}{k}$ converges by alternating series test and $(a_n)$ is non-negative, it follows that $(b_n)$ is bounded from below. Therefore $(b_n)$ converges, and so, $(a_n)$ converges as well.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    Thank you, that's neat! One might add that this argument always works for lower-bounded $(a_n)$ with $a_{n+1}le a_n+c_n$ for some summable $(c_n)$ by setting $b_n=a_n-sum_{k=1}^{n-1}c_k$.
    $endgroup$
    – Mars Plastic
    21 mins ago





















3












$begingroup$

As pointed out in the comments, this approach does not work. However, you can show that $(a_n)$ is a Cauchy sequence by using that
$$ sum_{n=1}^infty (a_{n+1}-a_n)lesum_{n=1}^infty (-1)^nfrac{1}{n}<infty.$$



EDIT: I am sorry, that was a bit too fast. Since the sequence is not assumed to be increasing, this alone does not help. I wanted to delete the answer, but it has already been accepted which apparently makes that impossible.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Okay kind of a dumb question here. I forgot how this would imply that the sequence is Cauchy. Do you have a source where I can see this for myself? I know I saw it somewhere but my memory isn't working for me. I think one of the key thing was that all bounded monotone sequences converge or something like that.
    $endgroup$
    – Darel
    1 hour ago












  • $begingroup$
    That doesn't show that it's Cauchy. Cauchy doesn't mean that the difference between successive terms goes to zero, it mean that $max_{n>N} |a_N-a_n|$ goes to zero as N goes to zero.
    $endgroup$
    – Acccumulation
    25 mins ago










  • $begingroup$
    1.) I did not claim that, not even in my original answer. 2.) Read my edit.
    $endgroup$
    – Mars Plastic
    20 mins ago










  • $begingroup$
    I unlocked it and gave appropriate credit to the clearer answer.
    $endgroup$
    – Darel
    2 mins ago



















0












$begingroup$

Use $$sum_{k=1}^n(a_{k+1}-a_k)=a_{n+1}-a_1leqsum_{k=1}^nfrac{(-1)^k}{k}rightarrow -ln2$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Be careful! The assumption is only an inequality.
    $endgroup$
    – Mars Plastic
    1 hour ago






  • 1




    $begingroup$
    @Mars Plastic I see. It was typo.
    $endgroup$
    – Michael Rozenberg
    1 hour ago










  • $begingroup$
    That shows that $(a_n)$ is bounded above, but why is it convergent?
    $endgroup$
    – Martin R
    30 mins ago











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

Consider $b_n = a_n + sum_{k=1}^{n-1} frac{(-1)^{k-1}}{k}$. Then



$$ b_{n+1}
= a_{n+1} + sum_{k=1}^{n} frac{(-1)^{k-1}}{k}
leq a_n + frac{(-1)^n}{n} + sum_{k=1}^{n} frac{(-1)^{k-1}}{k}
= b_n, $$



which shows that $(b_n)$ is non-increasing. Moreover, since $sum_{k=1}^{infty} frac{(-1)^{k-1}}{k}$ converges by alternating series test and $(a_n)$ is non-negative, it follows that $(b_n)$ is bounded from below. Therefore $(b_n)$ converges, and so, $(a_n)$ converges as well.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    Thank you, that's neat! One might add that this argument always works for lower-bounded $(a_n)$ with $a_{n+1}le a_n+c_n$ for some summable $(c_n)$ by setting $b_n=a_n-sum_{k=1}^{n-1}c_k$.
    $endgroup$
    – Mars Plastic
    21 mins ago


















3












$begingroup$

Consider $b_n = a_n + sum_{k=1}^{n-1} frac{(-1)^{k-1}}{k}$. Then



$$ b_{n+1}
= a_{n+1} + sum_{k=1}^{n} frac{(-1)^{k-1}}{k}
leq a_n + frac{(-1)^n}{n} + sum_{k=1}^{n} frac{(-1)^{k-1}}{k}
= b_n, $$



which shows that $(b_n)$ is non-increasing. Moreover, since $sum_{k=1}^{infty} frac{(-1)^{k-1}}{k}$ converges by alternating series test and $(a_n)$ is non-negative, it follows that $(b_n)$ is bounded from below. Therefore $(b_n)$ converges, and so, $(a_n)$ converges as well.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    Thank you, that's neat! One might add that this argument always works for lower-bounded $(a_n)$ with $a_{n+1}le a_n+c_n$ for some summable $(c_n)$ by setting $b_n=a_n-sum_{k=1}^{n-1}c_k$.
    $endgroup$
    – Mars Plastic
    21 mins ago
















3












3








3





$begingroup$

Consider $b_n = a_n + sum_{k=1}^{n-1} frac{(-1)^{k-1}}{k}$. Then



$$ b_{n+1}
= a_{n+1} + sum_{k=1}^{n} frac{(-1)^{k-1}}{k}
leq a_n + frac{(-1)^n}{n} + sum_{k=1}^{n} frac{(-1)^{k-1}}{k}
= b_n, $$



which shows that $(b_n)$ is non-increasing. Moreover, since $sum_{k=1}^{infty} frac{(-1)^{k-1}}{k}$ converges by alternating series test and $(a_n)$ is non-negative, it follows that $(b_n)$ is bounded from below. Therefore $(b_n)$ converges, and so, $(a_n)$ converges as well.






share|cite|improve this answer









$endgroup$



Consider $b_n = a_n + sum_{k=1}^{n-1} frac{(-1)^{k-1}}{k}$. Then



$$ b_{n+1}
= a_{n+1} + sum_{k=1}^{n} frac{(-1)^{k-1}}{k}
leq a_n + frac{(-1)^n}{n} + sum_{k=1}^{n} frac{(-1)^{k-1}}{k}
= b_n, $$



which shows that $(b_n)$ is non-increasing. Moreover, since $sum_{k=1}^{infty} frac{(-1)^{k-1}}{k}$ converges by alternating series test and $(a_n)$ is non-negative, it follows that $(b_n)$ is bounded from below. Therefore $(b_n)$ converges, and so, $(a_n)$ converges as well.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 38 mins ago









Sangchul LeeSangchul Lee

95k12170276




95k12170276








  • 2




    $begingroup$
    Thank you, that's neat! One might add that this argument always works for lower-bounded $(a_n)$ with $a_{n+1}le a_n+c_n$ for some summable $(c_n)$ by setting $b_n=a_n-sum_{k=1}^{n-1}c_k$.
    $endgroup$
    – Mars Plastic
    21 mins ago
















  • 2




    $begingroup$
    Thank you, that's neat! One might add that this argument always works for lower-bounded $(a_n)$ with $a_{n+1}le a_n+c_n$ for some summable $(c_n)$ by setting $b_n=a_n-sum_{k=1}^{n-1}c_k$.
    $endgroup$
    – Mars Plastic
    21 mins ago










2




2




$begingroup$
Thank you, that's neat! One might add that this argument always works for lower-bounded $(a_n)$ with $a_{n+1}le a_n+c_n$ for some summable $(c_n)$ by setting $b_n=a_n-sum_{k=1}^{n-1}c_k$.
$endgroup$
– Mars Plastic
21 mins ago






$begingroup$
Thank you, that's neat! One might add that this argument always works for lower-bounded $(a_n)$ with $a_{n+1}le a_n+c_n$ for some summable $(c_n)$ by setting $b_n=a_n-sum_{k=1}^{n-1}c_k$.
$endgroup$
– Mars Plastic
21 mins ago













3












$begingroup$

As pointed out in the comments, this approach does not work. However, you can show that $(a_n)$ is a Cauchy sequence by using that
$$ sum_{n=1}^infty (a_{n+1}-a_n)lesum_{n=1}^infty (-1)^nfrac{1}{n}<infty.$$



EDIT: I am sorry, that was a bit too fast. Since the sequence is not assumed to be increasing, this alone does not help. I wanted to delete the answer, but it has already been accepted which apparently makes that impossible.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Okay kind of a dumb question here. I forgot how this would imply that the sequence is Cauchy. Do you have a source where I can see this for myself? I know I saw it somewhere but my memory isn't working for me. I think one of the key thing was that all bounded monotone sequences converge or something like that.
    $endgroup$
    – Darel
    1 hour ago












  • $begingroup$
    That doesn't show that it's Cauchy. Cauchy doesn't mean that the difference between successive terms goes to zero, it mean that $max_{n>N} |a_N-a_n|$ goes to zero as N goes to zero.
    $endgroup$
    – Acccumulation
    25 mins ago










  • $begingroup$
    1.) I did not claim that, not even in my original answer. 2.) Read my edit.
    $endgroup$
    – Mars Plastic
    20 mins ago










  • $begingroup$
    I unlocked it and gave appropriate credit to the clearer answer.
    $endgroup$
    – Darel
    2 mins ago
















3












$begingroup$

As pointed out in the comments, this approach does not work. However, you can show that $(a_n)$ is a Cauchy sequence by using that
$$ sum_{n=1}^infty (a_{n+1}-a_n)lesum_{n=1}^infty (-1)^nfrac{1}{n}<infty.$$



EDIT: I am sorry, that was a bit too fast. Since the sequence is not assumed to be increasing, this alone does not help. I wanted to delete the answer, but it has already been accepted which apparently makes that impossible.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Okay kind of a dumb question here. I forgot how this would imply that the sequence is Cauchy. Do you have a source where I can see this for myself? I know I saw it somewhere but my memory isn't working for me. I think one of the key thing was that all bounded monotone sequences converge or something like that.
    $endgroup$
    – Darel
    1 hour ago












  • $begingroup$
    That doesn't show that it's Cauchy. Cauchy doesn't mean that the difference between successive terms goes to zero, it mean that $max_{n>N} |a_N-a_n|$ goes to zero as N goes to zero.
    $endgroup$
    – Acccumulation
    25 mins ago










  • $begingroup$
    1.) I did not claim that, not even in my original answer. 2.) Read my edit.
    $endgroup$
    – Mars Plastic
    20 mins ago










  • $begingroup$
    I unlocked it and gave appropriate credit to the clearer answer.
    $endgroup$
    – Darel
    2 mins ago














3












3








3





$begingroup$

As pointed out in the comments, this approach does not work. However, you can show that $(a_n)$ is a Cauchy sequence by using that
$$ sum_{n=1}^infty (a_{n+1}-a_n)lesum_{n=1}^infty (-1)^nfrac{1}{n}<infty.$$



EDIT: I am sorry, that was a bit too fast. Since the sequence is not assumed to be increasing, this alone does not help. I wanted to delete the answer, but it has already been accepted which apparently makes that impossible.






share|cite|improve this answer











$endgroup$



As pointed out in the comments, this approach does not work. However, you can show that $(a_n)$ is a Cauchy sequence by using that
$$ sum_{n=1}^infty (a_{n+1}-a_n)lesum_{n=1}^infty (-1)^nfrac{1}{n}<infty.$$



EDIT: I am sorry, that was a bit too fast. Since the sequence is not assumed to be increasing, this alone does not help. I wanted to delete the answer, but it has already been accepted which apparently makes that impossible.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 52 mins ago

























answered 2 hours ago









Mars PlasticMars Plastic

1,07918




1,07918












  • $begingroup$
    Okay kind of a dumb question here. I forgot how this would imply that the sequence is Cauchy. Do you have a source where I can see this for myself? I know I saw it somewhere but my memory isn't working for me. I think one of the key thing was that all bounded monotone sequences converge or something like that.
    $endgroup$
    – Darel
    1 hour ago












  • $begingroup$
    That doesn't show that it's Cauchy. Cauchy doesn't mean that the difference between successive terms goes to zero, it mean that $max_{n>N} |a_N-a_n|$ goes to zero as N goes to zero.
    $endgroup$
    – Acccumulation
    25 mins ago










  • $begingroup$
    1.) I did not claim that, not even in my original answer. 2.) Read my edit.
    $endgroup$
    – Mars Plastic
    20 mins ago










  • $begingroup$
    I unlocked it and gave appropriate credit to the clearer answer.
    $endgroup$
    – Darel
    2 mins ago


















  • $begingroup$
    Okay kind of a dumb question here. I forgot how this would imply that the sequence is Cauchy. Do you have a source where I can see this for myself? I know I saw it somewhere but my memory isn't working for me. I think one of the key thing was that all bounded monotone sequences converge or something like that.
    $endgroup$
    – Darel
    1 hour ago












  • $begingroup$
    That doesn't show that it's Cauchy. Cauchy doesn't mean that the difference between successive terms goes to zero, it mean that $max_{n>N} |a_N-a_n|$ goes to zero as N goes to zero.
    $endgroup$
    – Acccumulation
    25 mins ago










  • $begingroup$
    1.) I did not claim that, not even in my original answer. 2.) Read my edit.
    $endgroup$
    – Mars Plastic
    20 mins ago










  • $begingroup$
    I unlocked it and gave appropriate credit to the clearer answer.
    $endgroup$
    – Darel
    2 mins ago
















$begingroup$
Okay kind of a dumb question here. I forgot how this would imply that the sequence is Cauchy. Do you have a source where I can see this for myself? I know I saw it somewhere but my memory isn't working for me. I think one of the key thing was that all bounded monotone sequences converge or something like that.
$endgroup$
– Darel
1 hour ago






$begingroup$
Okay kind of a dumb question here. I forgot how this would imply that the sequence is Cauchy. Do you have a source where I can see this for myself? I know I saw it somewhere but my memory isn't working for me. I think one of the key thing was that all bounded monotone sequences converge or something like that.
$endgroup$
– Darel
1 hour ago














$begingroup$
That doesn't show that it's Cauchy. Cauchy doesn't mean that the difference between successive terms goes to zero, it mean that $max_{n>N} |a_N-a_n|$ goes to zero as N goes to zero.
$endgroup$
– Acccumulation
25 mins ago




$begingroup$
That doesn't show that it's Cauchy. Cauchy doesn't mean that the difference between successive terms goes to zero, it mean that $max_{n>N} |a_N-a_n|$ goes to zero as N goes to zero.
$endgroup$
– Acccumulation
25 mins ago












$begingroup$
1.) I did not claim that, not even in my original answer. 2.) Read my edit.
$endgroup$
– Mars Plastic
20 mins ago




$begingroup$
1.) I did not claim that, not even in my original answer. 2.) Read my edit.
$endgroup$
– Mars Plastic
20 mins ago












$begingroup$
I unlocked it and gave appropriate credit to the clearer answer.
$endgroup$
– Darel
2 mins ago




$begingroup$
I unlocked it and gave appropriate credit to the clearer answer.
$endgroup$
– Darel
2 mins ago











0












$begingroup$

Use $$sum_{k=1}^n(a_{k+1}-a_k)=a_{n+1}-a_1leqsum_{k=1}^nfrac{(-1)^k}{k}rightarrow -ln2$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Be careful! The assumption is only an inequality.
    $endgroup$
    – Mars Plastic
    1 hour ago






  • 1




    $begingroup$
    @Mars Plastic I see. It was typo.
    $endgroup$
    – Michael Rozenberg
    1 hour ago










  • $begingroup$
    That shows that $(a_n)$ is bounded above, but why is it convergent?
    $endgroup$
    – Martin R
    30 mins ago
















0












$begingroup$

Use $$sum_{k=1}^n(a_{k+1}-a_k)=a_{n+1}-a_1leqsum_{k=1}^nfrac{(-1)^k}{k}rightarrow -ln2$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Be careful! The assumption is only an inequality.
    $endgroup$
    – Mars Plastic
    1 hour ago






  • 1




    $begingroup$
    @Mars Plastic I see. It was typo.
    $endgroup$
    – Michael Rozenberg
    1 hour ago










  • $begingroup$
    That shows that $(a_n)$ is bounded above, but why is it convergent?
    $endgroup$
    – Martin R
    30 mins ago














0












0








0





$begingroup$

Use $$sum_{k=1}^n(a_{k+1}-a_k)=a_{n+1}-a_1leqsum_{k=1}^nfrac{(-1)^k}{k}rightarrow -ln2$$






share|cite|improve this answer











$endgroup$



Use $$sum_{k=1}^n(a_{k+1}-a_k)=a_{n+1}-a_1leqsum_{k=1}^nfrac{(-1)^k}{k}rightarrow -ln2$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 1 hour ago

























answered 2 hours ago









Michael RozenbergMichael Rozenberg

106k1893198




106k1893198








  • 1




    $begingroup$
    Be careful! The assumption is only an inequality.
    $endgroup$
    – Mars Plastic
    1 hour ago






  • 1




    $begingroup$
    @Mars Plastic I see. It was typo.
    $endgroup$
    – Michael Rozenberg
    1 hour ago










  • $begingroup$
    That shows that $(a_n)$ is bounded above, but why is it convergent?
    $endgroup$
    – Martin R
    30 mins ago














  • 1




    $begingroup$
    Be careful! The assumption is only an inequality.
    $endgroup$
    – Mars Plastic
    1 hour ago






  • 1




    $begingroup$
    @Mars Plastic I see. It was typo.
    $endgroup$
    – Michael Rozenberg
    1 hour ago










  • $begingroup$
    That shows that $(a_n)$ is bounded above, but why is it convergent?
    $endgroup$
    – Martin R
    30 mins ago








1




1




$begingroup$
Be careful! The assumption is only an inequality.
$endgroup$
– Mars Plastic
1 hour ago




$begingroup$
Be careful! The assumption is only an inequality.
$endgroup$
– Mars Plastic
1 hour ago




1




1




$begingroup$
@Mars Plastic I see. It was typo.
$endgroup$
– Michael Rozenberg
1 hour ago




$begingroup$
@Mars Plastic I see. It was typo.
$endgroup$
– Michael Rozenberg
1 hour ago












$begingroup$
That shows that $(a_n)$ is bounded above, but why is it convergent?
$endgroup$
– Martin R
30 mins ago




$begingroup$
That shows that $(a_n)$ is bounded above, but why is it convergent?
$endgroup$
– Martin R
30 mins ago


















draft saved

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