Show that the following sequence converges. Please Critique my proof.Prove that a sequence converges to a...
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Show that the following sequence converges. Please Critique my proof.
Prove that a sequence converges to a finite limit iff lim inf equals lim supShow the convergence of sequenceWhat is wrong with the following proof?Simple proof that this sequence converges [verification]Proof that the sequence $a_n=frac{3n+2}{n^2+1}$ converges using the Epsilon N proofQuestion about the proof that the sequence ${a_{j}cdot b_{j}}$ converges to $alpha beta $show that if a subsequence of a cauchy sequence converges, then the whole sequence convergesProof that bounded growth of a sequence implies convergenceShow that the sequence $a_n=frac{cos(n^2+n)}{n^2}$ converges to $0$.Proof that the sequence $left{frac{5n^2-6}{2n^3-7n}right}$ converges to $0$
$begingroup$
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
asked 11 hours ago
DarelDarel
1199
$endgroup$
add a comment |
$begingroup$
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
asked 11 hours ago
DarelDarel
1199
$endgroup$
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
11 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
10 hours ago
add a comment |
$begingroup$
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
asked 11 hours ago
DarelDarel
1199
$endgroup$
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
asked 11 hours ago
DarelDarel
1199
asked 11 hours ago
DarelDarel
1199
edited 7 hours ago
Darel
asked 11 hours ago
DarelDarel
1199
asked 11 hours ago
DarelDarel
1199
asked 11 hours ago
DarelDarel
1199
1199
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
11 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
10 hours ago
add a comment |
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
11 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
10 hours 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
11 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
11 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
10 hours 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
10 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered 10 hours ago
Sangchul LeeSangchul Lee
95k12170276
$endgroup$
3
$begingroup$
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
9 hours ago
$begingroup$
@MarsPlastic The argument works even if $(a_n)$ is not bounded below, in that case $a_n to -infty$ follows.
$endgroup$
– Martin R
1 hour ago
$begingroup$
@MarsPlastic: Btw, thank you for pointing out the flaw in my answer. I was able to fix that, but this answer is so elegant and much simpler, that I deleted mine again.
$endgroup$
– Martin R
47 mins ago
$begingroup$
This is really an elegant way. Unfortunately I just got it done by brute force.
$endgroup$
– Falrach
14 mins ago
add a comment |
$begingroup$
Define $b_k := a_{2k+1}$. Then
$$b_k leq a_{2k} + (-1)^{2k}frac{1}{2k} leq b_{k-1} + (frac{1}{2k} - frac{1}{2k-1}) leq b_{k-1}$$
Since $b_k$ is non-negative and non-increasing: $b_k to b$.
Suppose $a_n nrightarrow b$. Then there exists an $varepsilon > 0 $ s.t. for infinitely many $n$ holds $|a_{2n} - b| > varepsilon$.
Assume that $|a_{2m+1}-a_m| > frac{varepsilon}{2}$ for infinitely many $m$. Then, since $a_{2m+1}- a_m leq frac{1}{2m}$ we have that
begin{align}
a_{2m+1} - a_m < - frac{varepsilon}{2}
end{align}
for infinitely many $m$. Let $M := {m geq 1 : a_{2m+1} - a_m < - frac{varepsilon}{2} text{ is fulfilled for } m }$
begin{align*}
d_m := 1_M (m)
end{align*}
This implies
begin{align*}
0 leq a_{2m+1} = a_1 + sum_{k=1}^{2m} (a_{k+1} - a_k ) = a_1 + sum_{k=1}^m (a_{2k+1} - a_{2k}) + sum_{k=1}^m (a_{2k} - {a_{2k-1}}) \
leq a_1 + sum_{k=1}^m (-1)^{2k} frac{1}{2k}- frac{varepsilon}{2} d_k + sum_{k=1}^m (-1)^{2k-1}frac{1}{2k-1} to a_1 - sum_{k=1}^infty frac{varepsilon}{2} d_k + sum_{i=1}^infty (-1)^i frac{1}{i} = - infty
end{align*}
since $|M| = infty$ and the last series converges. This is a contradiction.
Therefore we have that there exists $Kgeq 1$ s.t. for all $kgeq K$ it holds: $|a_{2k+1} - a_k| leq frac{varepsilon}{2}$. We can conclude that
begin{align*}
|a{2n+1} - b| geq |a_{2n} - b| - |a_{2n+1} - a_n| geq varepsilon - frac{varepsilon}{2} = frac{varepsilon}{2}
end{align*}
for infinitely $n geq K$. Contradiction. Thus $a_n to b$.
answered 7 hours ago
FalrachFalrach
1,676224
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered 10 hours ago
Sangchul LeeSangchul Lee
95k12170276
$endgroup$
3
$begingroup$
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
9 hours ago
$begingroup$
@MarsPlastic The argument works even if $(a_n)$ is not bounded below, in that case $a_n to -infty$ follows.
$endgroup$
– Martin R
1 hour ago
$begingroup$
@MarsPlastic: Btw, thank you for pointing out the flaw in my answer. I was able to fix that, but this answer is so elegant and much simpler, that I deleted mine again.
$endgroup$
– Martin R
47 mins ago
$begingroup$
This is really an elegant way. Unfortunately I just got it done by brute force.
$endgroup$
– Falrach
14 mins ago
add a comment |
$begingroup$
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.
answered 10 hours ago
Sangchul LeeSangchul Lee
95k12170276
$endgroup$
3
$begingroup$
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
9 hours ago
$begingroup$
@MarsPlastic The argument works even if $(a_n)$ is not bounded below, in that case $a_n to -infty$ follows.
$endgroup$
– Martin R
1 hour ago
$begingroup$
@MarsPlastic: Btw, thank you for pointing out the flaw in my answer. I was able to fix that, but this answer is so elegant and much simpler, that I deleted mine again.
$endgroup$
– Martin R
47 mins ago
$begingroup$
This is really an elegant way. Unfortunately I just got it done by brute force.
$endgroup$
– Falrach
14 mins ago
add a comment |
$begingroup$
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.
answered 10 hours ago
Sangchul LeeSangchul Lee
95k12170276
$endgroup$
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.
answered 10 hours ago
Sangchul LeeSangchul Lee
95k12170276
answered 10 hours ago
Sangchul LeeSangchul Lee
95k12170276
answered 10 hours ago
Sangchul LeeSangchul Lee
95k12170276
answered 10 hours ago
Sangchul LeeSangchul Lee
95k12170276
95k12170276
3
$begingroup$
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
9 hours ago
$begingroup$
@MarsPlastic The argument works even if $(a_n)$ is not bounded below, in that case $a_n to -infty$ follows.
$endgroup$
– Martin R
1 hour ago
$begingroup$
@MarsPlastic: Btw, thank you for pointing out the flaw in my answer. I was able to fix that, but this answer is so elegant and much simpler, that I deleted mine again.
$endgroup$
– Martin R
47 mins ago
$begingroup$
This is really an elegant way. Unfortunately I just got it done by brute force.
$endgroup$
– Falrach
14 mins ago
add a comment |
3
$begingroup$
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
9 hours ago
$begingroup$
@MarsPlastic The argument works even if $(a_n)$ is not bounded below, in that case $a_n to -infty$ follows.
$endgroup$
– Martin R
1 hour ago
$begingroup$
@MarsPlastic: Btw, thank you for pointing out the flaw in my answer. I was able to fix that, but this answer is so elegant and much simpler, that I deleted mine again.
$endgroup$
– Martin R
47 mins ago
$begingroup$
This is really an elegant way. Unfortunately I just got it done by brute force.
$endgroup$
– Falrach
14 mins ago
3
3
$begingroup$
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
9 hours 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
9 hours ago
$begingroup$
@MarsPlastic The argument works even if $(a_n)$ is not bounded below, in that case $a_n to -infty$ follows.
$endgroup$
– Martin R
1 hour ago
$begingroup$
@MarsPlastic The argument works even if $(a_n)$ is not bounded below, in that case $a_n to -infty$ follows.
$endgroup$
– Martin R
1 hour ago
$begingroup$
@MarsPlastic: Btw, thank you for pointing out the flaw in my answer. I was able to fix that, but this answer is so elegant and much simpler, that I deleted mine again.
$endgroup$
– Martin R
47 mins ago
$begingroup$
@MarsPlastic: Btw, thank you for pointing out the flaw in my answer. I was able to fix that, but this answer is so elegant and much simpler, that I deleted mine again.
$endgroup$
– Martin R
47 mins ago
$begingroup$
This is really an elegant way. Unfortunately I just got it done by brute force.
$endgroup$
– Falrach
14 mins ago
$begingroup$
This is really an elegant way. Unfortunately I just got it done by brute force.
$endgroup$
– Falrach
14 mins ago
add a comment |
$begingroup$
Define $b_k := a_{2k+1}$. Then
$$b_k leq a_{2k} + (-1)^{2k}frac{1}{2k} leq b_{k-1} + (frac{1}{2k} - frac{1}{2k-1}) leq b_{k-1}$$
Since $b_k$ is non-negative and non-increasing: $b_k to b$.
Suppose $a_n nrightarrow b$. Then there exists an $varepsilon > 0 $ s.t. for infinitely many $n$ holds $|a_{2n} - b| > varepsilon$.
Assume that $|a_{2m+1}-a_m| > frac{varepsilon}{2}$ for infinitely many $m$. Then, since $a_{2m+1}- a_m leq frac{1}{2m}$ we have that
begin{align}
a_{2m+1} - a_m < - frac{varepsilon}{2}
end{align}
for infinitely many $m$. Let $M := {m geq 1 : a_{2m+1} - a_m < - frac{varepsilon}{2} text{ is fulfilled for } m }$
begin{align*}
d_m := 1_M (m)
end{align*}
This implies
begin{align*}
0 leq a_{2m+1} = a_1 + sum_{k=1}^{2m} (a_{k+1} - a_k ) = a_1 + sum_{k=1}^m (a_{2k+1} - a_{2k}) + sum_{k=1}^m (a_{2k} - {a_{2k-1}}) \
leq a_1 + sum_{k=1}^m (-1)^{2k} frac{1}{2k}- frac{varepsilon}{2} d_k + sum_{k=1}^m (-1)^{2k-1}frac{1}{2k-1} to a_1 - sum_{k=1}^infty frac{varepsilon}{2} d_k + sum_{i=1}^infty (-1)^i frac{1}{i} = - infty
end{align*}
since $|M| = infty$ and the last series converges. This is a contradiction.
Therefore we have that there exists $Kgeq 1$ s.t. for all $kgeq K$ it holds: $|a_{2k+1} - a_k| leq frac{varepsilon}{2}$. We can conclude that
begin{align*}
|a{2n+1} - b| geq |a_{2n} - b| - |a_{2n+1} - a_n| geq varepsilon - frac{varepsilon}{2} = frac{varepsilon}{2}
end{align*}
for infinitely $n geq K$. Contradiction. Thus $a_n to b$.
answered 7 hours ago
FalrachFalrach
1,676224
$endgroup$
add a comment |
$begingroup$
Define $b_k := a_{2k+1}$. Then
$$b_k leq a_{2k} + (-1)^{2k}frac{1}{2k} leq b_{k-1} + (frac{1}{2k} - frac{1}{2k-1}) leq b_{k-1}$$
Since $b_k$ is non-negative and non-increasing: $b_k to b$.
Suppose $a_n nrightarrow b$. Then there exists an $varepsilon > 0 $ s.t. for infinitely many $n$ holds $|a_{2n} - b| > varepsilon$.
Assume that $|a_{2m+1}-a_m| > frac{varepsilon}{2}$ for infinitely many $m$. Then, since $a_{2m+1}- a_m leq frac{1}{2m}$ we have that
begin{align}
a_{2m+1} - a_m < - frac{varepsilon}{2}
end{align}
for infinitely many $m$. Let $M := {m geq 1 : a_{2m+1} - a_m < - frac{varepsilon}{2} text{ is fulfilled for } m }$
begin{align*}
d_m := 1_M (m)
end{align*}
This implies
begin{align*}
0 leq a_{2m+1} = a_1 + sum_{k=1}^{2m} (a_{k+1} - a_k ) = a_1 + sum_{k=1}^m (a_{2k+1} - a_{2k}) + sum_{k=1}^m (a_{2k} - {a_{2k-1}}) \
leq a_1 + sum_{k=1}^m (-1)^{2k} frac{1}{2k}- frac{varepsilon}{2} d_k + sum_{k=1}^m (-1)^{2k-1}frac{1}{2k-1} to a_1 - sum_{k=1}^infty frac{varepsilon}{2} d_k + sum_{i=1}^infty (-1)^i frac{1}{i} = - infty
end{align*}
since $|M| = infty$ and the last series converges. This is a contradiction.
Therefore we have that there exists $Kgeq 1$ s.t. for all $kgeq K$ it holds: $|a_{2k+1} - a_k| leq frac{varepsilon}{2}$. We can conclude that
begin{align*}
|a{2n+1} - b| geq |a_{2n} - b| - |a_{2n+1} - a_n| geq varepsilon - frac{varepsilon}{2} = frac{varepsilon}{2}
end{align*}
for infinitely $n geq K$. Contradiction. Thus $a_n to b$.
answered 7 hours ago
FalrachFalrach
1,676224
$endgroup$
add a comment |
$begingroup$
Define $b_k := a_{2k+1}$. Then
$$b_k leq a_{2k} + (-1)^{2k}frac{1}{2k} leq b_{k-1} + (frac{1}{2k} - frac{1}{2k-1}) leq b_{k-1}$$
Since $b_k$ is non-negative and non-increasing: $b_k to b$.
Suppose $a_n nrightarrow b$. Then there exists an $varepsilon > 0 $ s.t. for infinitely many $n$ holds $|a_{2n} - b| > varepsilon$.
Assume that $|a_{2m+1}-a_m| > frac{varepsilon}{2}$ for infinitely many $m$. Then, since $a_{2m+1}- a_m leq frac{1}{2m}$ we have that
begin{align}
a_{2m+1} - a_m < - frac{varepsilon}{2}
end{align}
for infinitely many $m$. Let $M := {m geq 1 : a_{2m+1} - a_m < - frac{varepsilon}{2} text{ is fulfilled for } m }$
begin{align*}
d_m := 1_M (m)
end{align*}
This implies
begin{align*}
0 leq a_{2m+1} = a_1 + sum_{k=1}^{2m} (a_{k+1} - a_k ) = a_1 + sum_{k=1}^m (a_{2k+1} - a_{2k}) + sum_{k=1}^m (a_{2k} - {a_{2k-1}}) \
leq a_1 + sum_{k=1}^m (-1)^{2k} frac{1}{2k}- frac{varepsilon}{2} d_k + sum_{k=1}^m (-1)^{2k-1}frac{1}{2k-1} to a_1 - sum_{k=1}^infty frac{varepsilon}{2} d_k + sum_{i=1}^infty (-1)^i frac{1}{i} = - infty
end{align*}
since $|M| = infty$ and the last series converges. This is a contradiction.
Therefore we have that there exists $Kgeq 1$ s.t. for all $kgeq K$ it holds: $|a_{2k+1} - a_k| leq frac{varepsilon}{2}$. We can conclude that
begin{align*}
|a{2n+1} - b| geq |a_{2n} - b| - |a_{2n+1} - a_n| geq varepsilon - frac{varepsilon}{2} = frac{varepsilon}{2}
end{align*}
for infinitely $n geq K$. Contradiction. Thus $a_n to b$.
answered 7 hours ago
FalrachFalrach
1,676224
$endgroup$
Define $b_k := a_{2k+1}$. Then
$$b_k leq a_{2k} + (-1)^{2k}frac{1}{2k} leq b_{k-1} + (frac{1}{2k} - frac{1}{2k-1}) leq b_{k-1}$$
Since $b_k$ is non-negative and non-increasing: $b_k to b$.
Suppose $a_n nrightarrow b$. Then there exists an $varepsilon > 0 $ s.t. for infinitely many $n$ holds $|a_{2n} - b| > varepsilon$.
Assume that $|a_{2m+1}-a_m| > frac{varepsilon}{2}$ for infinitely many $m$. Then, since $a_{2m+1}- a_m leq frac{1}{2m}$ we have that
begin{align}
a_{2m+1} - a_m < - frac{varepsilon}{2}
end{align}
for infinitely many $m$. Let $M := {m geq 1 : a_{2m+1} - a_m < - frac{varepsilon}{2} text{ is fulfilled for } m }$
begin{align*}
d_m := 1_M (m)
end{align*}
This implies
begin{align*}
0 leq a_{2m+1} = a_1 + sum_{k=1}^{2m} (a_{k+1} - a_k ) = a_1 + sum_{k=1}^m (a_{2k+1} - a_{2k}) + sum_{k=1}^m (a_{2k} - {a_{2k-1}}) \
leq a_1 + sum_{k=1}^m (-1)^{2k} frac{1}{2k}- frac{varepsilon}{2} d_k + sum_{k=1}^m (-1)^{2k-1}frac{1}{2k-1} to a_1 - sum_{k=1}^infty frac{varepsilon}{2} d_k + sum_{i=1}^infty (-1)^i frac{1}{i} = - infty
end{align*}
since $|M| = infty$ and the last series converges. This is a contradiction.
Therefore we have that there exists $Kgeq 1$ s.t. for all $kgeq K$ it holds: $|a_{2k+1} - a_k| leq frac{varepsilon}{2}$. We can conclude that
begin{align*}
|a{2n+1} - b| geq |a_{2n} - b| - |a_{2n+1} - a_n| geq varepsilon - frac{varepsilon}{2} = frac{varepsilon}{2}
end{align*}
for infinitely $n geq K$. Contradiction. Thus $a_n to b$.
answered 7 hours ago
FalrachFalrach
1,676224
answered 7 hours ago
FalrachFalrach
1,676224
answered 7 hours ago
FalrachFalrach
1,676224
answered 7 hours ago
FalrachFalrach
1,676224
1,676224
add a comment |
add a comment |
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$begingroup$
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
11 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
10 hours ago