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 5 hours ago
DarelDarel
1149
$endgroup$
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 5 hours ago
DarelDarel
1149
$endgroup$
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
5 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
3 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 5 hours ago
DarelDarel
1149
$endgroup$
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 5 hours ago
DarelDarel
1149
asked 5 hours ago
DarelDarel
1149
edited 37 mins ago
Darel
asked 5 hours ago
DarelDarel
1149
asked 5 hours ago
DarelDarel
1149
asked 5 hours ago
DarelDarel
1149
1149
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
5 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
3 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
5 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
3 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
5 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
5 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
3 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
3 hours ago
add a comment |
3 Answers
3
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 3 hours ago
Sangchul LeeSangchul Lee
95k12170276
$endgroup$
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
3 hours 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 1 hour ago
FalrachFalrach
1,676224
$endgroup$
add a comment |
$begingroup$
Use $$sum_{k=1}^n(a_{k+1}-a_k)=a_{n+1}-a_1leqsum_{k=1}^nfrac{(-1)^k}{k}rightarrow -ln2$$
answered 4 hours ago
Michael RozenbergMichael Rozenberg
106k1893198
$endgroup$
1
$begingroup$
Be careful! The assumption is only an inequality.
$endgroup$
– Mars Plastic
4 hours ago
1
$begingroup$
@Mars Plastic I see. It was typo.
$endgroup$
– Michael Rozenberg
4 hours ago
1
$begingroup$
That shows that $(a_n)$ is bounded above, but why is it convergent?
$endgroup$
– Martin R
3 hours ago
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
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 3 hours ago
Sangchul LeeSangchul Lee
95k12170276
$endgroup$
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
3 hours 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 3 hours ago
Sangchul LeeSangchul Lee
95k12170276
$endgroup$
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
3 hours 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 3 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 3 hours ago
Sangchul LeeSangchul Lee
95k12170276
answered 3 hours ago
Sangchul LeeSangchul Lee
95k12170276
answered 3 hours ago
Sangchul LeeSangchul Lee
95k12170276
answered 3 hours 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
3 hours ago
add a comment |
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
3 hours 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
3 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
3 hours 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 1 hour 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 1 hour 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 1 hour 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 1 hour ago
FalrachFalrach
1,676224
answered 1 hour ago
FalrachFalrach
1,676224
answered 1 hour ago
FalrachFalrach
1,676224
answered 1 hour ago
FalrachFalrach
1,676224
1,676224
add a comment |
add a comment |
$begingroup$
Use $$sum_{k=1}^n(a_{k+1}-a_k)=a_{n+1}-a_1leqsum_{k=1}^nfrac{(-1)^k}{k}rightarrow -ln2$$
answered 4 hours ago
Michael RozenbergMichael Rozenberg
106k1893198
$endgroup$
1
$begingroup$
Be careful! The assumption is only an inequality.
$endgroup$
– Mars Plastic
4 hours ago
1
$begingroup$
@Mars Plastic I see. It was typo.
$endgroup$
– Michael Rozenberg
4 hours ago
1
$begingroup$
That shows that $(a_n)$ is bounded above, but why is it convergent?
$endgroup$
– Martin R
3 hours ago
add a comment |
$begingroup$
Use $$sum_{k=1}^n(a_{k+1}-a_k)=a_{n+1}-a_1leqsum_{k=1}^nfrac{(-1)^k}{k}rightarrow -ln2$$
answered 4 hours ago
Michael RozenbergMichael Rozenberg
106k1893198
$endgroup$
1
$begingroup$
Be careful! The assumption is only an inequality.
$endgroup$
– Mars Plastic
4 hours ago
1
$begingroup$
@Mars Plastic I see. It was typo.
$endgroup$
– Michael Rozenberg
4 hours ago
1
$begingroup$
That shows that $(a_n)$ is bounded above, but why is it convergent?
$endgroup$
– Martin R
3 hours ago
add a comment |
$begingroup$
Use $$sum_{k=1}^n(a_{k+1}-a_k)=a_{n+1}-a_1leqsum_{k=1}^nfrac{(-1)^k}{k}rightarrow -ln2$$
answered 4 hours ago
Michael RozenbergMichael Rozenberg
106k1893198
$endgroup$
Use $$sum_{k=1}^n(a_{k+1}-a_k)=a_{n+1}-a_1leqsum_{k=1}^nfrac{(-1)^k}{k}rightarrow -ln2$$
answered 4 hours ago
Michael RozenbergMichael Rozenberg
106k1893198
edited 4 hours ago
answered 4 hours ago
Michael RozenbergMichael Rozenberg
106k1893198
answered 4 hours ago
Michael RozenbergMichael Rozenberg
106k1893198
answered 4 hours ago
Michael RozenbergMichael Rozenberg
106k1893198
106k1893198
1
$begingroup$
Be careful! The assumption is only an inequality.
$endgroup$
– Mars Plastic
4 hours ago
1
$begingroup$
@Mars Plastic I see. It was typo.
$endgroup$
– Michael Rozenberg
4 hours ago
1
$begingroup$
That shows that $(a_n)$ is bounded above, but why is it convergent?
$endgroup$
– Martin R
3 hours ago
add a comment |
1
$begingroup$
Be careful! The assumption is only an inequality.
$endgroup$
– Mars Plastic
4 hours ago
1
$begingroup$
@Mars Plastic I see. It was typo.
$endgroup$
– Michael Rozenberg
4 hours ago
1
$begingroup$
That shows that $(a_n)$ is bounded above, but why is it convergent?
$endgroup$
– Martin R
3 hours ago
1
1
$begingroup$
Be careful! The assumption is only an inequality.
$endgroup$
– Mars Plastic
4 hours ago
$begingroup$
Be careful! The assumption is only an inequality.
$endgroup$
– Mars Plastic
4 hours ago
1
1
$begingroup$
@Mars Plastic I see. It was typo.
$endgroup$
– Michael Rozenberg
4 hours ago
$begingroup$
@Mars Plastic I see. It was typo.
$endgroup$
– Michael Rozenberg
4 hours ago
1
1
$begingroup$
That shows that $(a_n)$ is bounded above, but why is it convergent?
$endgroup$
– Martin R
3 hours ago
$begingroup$
That shows that $(a_n)$ is bounded above, but why is it convergent?
$endgroup$
– Martin R
3 hours ago
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
5 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
3 hours ago