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

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.

• Your proof is not correct. Your arguments would also work for $a_n = \sum_{i=1}^n \frac 1 i$, which does not converge. – Falrach Mar 1 at 21:13
• 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). – Maximilian Janisch Mar 1 at 22:49
• This is a duplicate, but I’m too lazy to find the original... – Shalop Mar 2 at 14:53
• @Shalop if you do find the original, then please tell me. – Darel Mar 2 at 16:35

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.
• 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 $b_n=a_n-\sum_{k=1}^{n-1}c_k$. – Mars Plastic Mar 1 at 23:04
• @MarsPlastic The argument works even if $(a_n)$ is not bounded below, in that case $a_n \to -\infty$ follows. – Martin R Mar 2 at 7:28
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 $$a_n \to b$$.