Yrurtj

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.

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.

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 $a_n to b$.

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