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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 $$ |a_{n+1}-a_n|\leq \left|\frac{(-1)^n}{n}\right|\leq\frac{1}{n} $$ and since it is known that $\frac{1}{n}\rightarrow 0$ as $n\rightarrow \infty$. 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.

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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 Mar 1 at 21:13
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    $\begingroup$ 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 Mar 1 at 22:49
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    $\begingroup$ This is a duplicate, but I’m too lazy to find the original... $\endgroup$ – Shalop Mar 2 at 14:53
  • $\begingroup$ @Shalop if you do find the original, then please tell me. $\endgroup$ – Darel Mar 2 at 16:35
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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.

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    $\begingroup$ 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 Mar 1 at 23:04
  • $\begingroup$ @MarsPlastic The argument works even if $(a_n)$ is not bounded below, in that case $a_n \to -\infty$ follows. $\endgroup$ – Martin R Mar 2 at 7:28
  • $\begingroup$ @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 Mar 2 at 8:04
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    $\begingroup$ This is really an elegant way. Unfortunately I just got it done by brute force. $\endgroup$ – Falrach Mar 2 at 8:36
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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 $K\geq 1$ s.t. for all $k\geq K$ 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 $a_n \to b$.

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