We have complex sequence $a_n$ such that $\displaystyle \sum_{n=1}^{\infty}a_n$ is convergent. Let $\sigma : \mathbb{N} \to \mathbb{N}$ be a bijection where we know that there exist $M \in \mathbb{N}$ such that for every $n \in \mathbb{N}$ we have $|\sigma(n)-n|<M$ .Prove that $\displaystyle \sum_{n=1}^{\infty}a_{\sigma(n)} $ is convergent.
My try :
Since $|\sigma(n)-n|<M$ so for some $k \in \mathbb{N}$ we have $\sigma(k) \le M+k$ because $\sigma$ is bijection then at the interval $[1, k+M]$ , $ \ \ \sigma$ takes all natural values from $[1, k]$. Let $\displaystyle S_k=\sum_{n=1}^{M+k}a_\sigma(n) $ then we have $\displaystyle S_k=\sum_{n=1}^{k}a_n + W$ where $W=a_{k+1}+...+a_M$ and here I have problem to show that $W \to 0$ (if it's true) and then as $k \to \infty$ we have $\displaystyle S_k \to \sum_{n=1}^{\infty}a_n$ so $\displaystyle \sum_{n=1}^{\infty}a_\sigma(n) \le \sum_{n=1}^{\infty}a_n$
Observe that
$$\sum_{n=1}^\infty a_n\;\;\text{converges}\;\iff\;\lim_{n\to\infty\,,\,m\to\infty}\sum_{k=n}^{n+m} a_k=0$$
The right hand condition is just Cauchy's Condition, which means the sequence of partial sum is a Cauchy sequence.
Thus, taking from where you were: using the above, show now that $\;W\xrightarrow[k\,,\,m\to\infty]{}0\;$