mjhjmtu

The problem is as follows:

Let $a_n$ be a sequence of nonnegative numbers such that
$$
a_n+1leq a_n+frac(-1)^nn.
$$
Show that $a_n$ converges.

My (wrong) proof:

Notice that
$$
|a_n+1-a_n|leq left|frac(-1)^nnright|leqfrac1n
$$
and since it is known that $frac1nrightarrow 0$ 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 $left|frac1nright|<epsilon$ as well?

Notes: Currently working on the proof.

Consider $b_n = a_n + sum_k=1^n-1 frac(-1)^k-1k$. Then

$$ b_n+1
= a_n+1 + sum_k=1^n frac(-1)^k-1k
leq a_n + frac(-1)^nn + sum_k=1^n frac(-1)^k-1k
= b_n, $$

which shows that $(b_n)$ is non-increasing. Moreover, since $sum_k=1^infty frac(-1)^k-1k$ 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
$$b_k leq a_2k + (-1)^2kfrac12k leq b_k-1 + (frac12k - frac12k-1) leq b_k-1$$
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 $|a_2m+1-a_m| > fracvarepsilon2$ for infinitely many $m$. Then, since $a_2m+1- a_m leq frac12m$ we have that
beginalign
a_2m+1 - a_m < - fracvarepsilon2
endalign
for infinitely many $m$. Let $M := m geq 1 : a_2m+1 - a_m < - fracvarepsilon2 text is fulfilled for m $
beginalign*
d_m := 1_M (m)
endalign*
This implies
beginalign*
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 frac12k- fracvarepsilon2 d_k + sum_k=1^m (-1)^2k-1frac12k-1 to a_1 - sum_k=1^infty fracvarepsilon2 d_k + sum_i=1^infty (-1)^i frac1i = - infty
endalign*
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: $|a_2k+1 - a_k| leq fracvarepsilon2$. We can conclude that
beginalign*
|a2n+1 - b| geq |a_2n - b| - |a_2n+1 - a_n| geq varepsilon - fracvarepsilon2 = fracvarepsilon2
endalign*
for infinitely $n geq K$. Contradiction. Thus $a_n to b$.