I encountered this problem while I was going through the book Analysis 1 by Otto Forster. I came up with a proof but I still am not sure myself whether the proof is valid or not. Sadly there is no given solution for this answer to which I can turn to. In the book, the set of $\mathbb{N}$ starts from $0$.
The problem is to prove that given $d\in \mathbb{N}$ and a bijection $\tau: \mathbb{N} \longrightarrow\mathbb{N} $ such that $|\tau(n)-n|\leq d$ for all $n\in \mathbb{N}$, then a series $\sum_{n=1}^{\infty} a_n $ converges, if and only if the series $\sum_{n=1}^{\infty} a_{\tau(n)} $ converges.
Proof. First I show that for any $m\geq d$ the following applies: $$\{a_0,a_1,...,a_{m-d} \}\subseteq \{a_{\tau(0)},a_{\tau(1)},...,a_{\tau(m)} \}$$since for any $x$, there exists a $y\in[x-d,x+d]$ so that $a_{\tau(y)}=a_x$.
This means that
$$\sum_{n=0}^{m} a_{\tau(n)} =\sum_{n=0}^{m-d} a_{n} + X_m $$
with
$$X_m = \sum_{n=0}^{d-1} a_{f_m(n)}$$
for any $m\geq d$ whereas $f$ is a function that maps the remaining a's with strictly increasing order of its indices, $$f_m(0)< f_m(2)< ... < f_m(d-1) $$
Then we construct "subsequences" of $(a_n)_{n\in \mathbb{N}}$ (the " " mark because $f$ may take on same values for finitely consecutive $j$, but the limit of the sequences will still be equal to $0$ since $m$ is increasing, and by definition of limit, we can take for any $\epsilon>0$ a $N\in\mathbb{N}$ to be $M$ the largest natural number such that $a_M=a_N$ with $N$ as in the definition limit of $a_n$, $$b_{j,k} = a_{f_j(k)}, \hspace{1cm} 0 \leq k \leq d-1. $$ So that
$$X_m=b_{m,0}+b_{m,1}+...+b_{m,d-1}$$
Since the series $\sum_{n=1}^{\infty} a_n $ converges, $\lim_{n\to\infty} a(n) = 0$, so is it also with every of its subsequences. Hence we have:
$$\lim_{m\to\infty}\sum_{n=0}^{m} a_{\tau(n)} =\lim_{m\to\infty}\sum_{n=0}^{m-d} a_{n} + \lim_{m\to\infty}X_m=\sum_{n=0}^{\infty} a_{n}+\lim_{m\to\infty}b_{m,0}+\lim_{m\to\infty}b_{m,1}+...+\lim_{m\to\infty}b_{m,d-1} =\sum_{n=0}^{\infty}a_{n} .$$
The proof of the other direction follows by the use of above in reversing the role. $q.e.d.$
Another known approach to this proof would be by using the definition of limits and with the use of $\frac{\epsilon}{d+1}$, so that with the triangle inequality, $|X_m|\leq \frac{d* \epsilon}{d+1}$ so that they sum up to $\epsilon$, which approach is actually the better one?
Any feedback would be greatly appreciated, if the proof above is not valid, please point out the errors and how to fix them, thank you, or if there is no obvious way to fix it,then another idea of proof is also appreciated. But if the proof is already correct, please suggest a way to write a better one, perhaps with excluding some unnecessary explanations or other faster/more obvious approach, thank you very much!