Converging series and bounds on partial sum

Question

I am currently working on a proof:

Let $(a_n) \subset \mathbb{R}$ be a sequence, and $s_k := \sum^{k}_{n=1} a_n$, show that: If, for all $k \in \mathbb{N} :|s_{k+1} - s_k | < \frac{1}{2^k} \implies \sum^{\infty}_{n=1} a_n$ converges

Now my first idea was that all terms of the sum except $a_{k+1}$ cancel out so, for every k the sequence $(a_n)$ seems to be bounded by $\frac{1}{2^k}$, is that correct?

How could I argue that this sequence is monoton?

Answer 1

For all $k>1 $

$$|s_{k}-s_{k-1}|=|a_k|<\frac {1}{2^{k-1}} $$

and comparison test.

Answer 2

Not necessarily monotone but it is some sense of Cauchy:

For $p=1,2,...$, \begin{align*} |s_{n+p}-s_{n}|&\leq|s_{n+p}-s_{n+p-1}+\cdots+s_{n+1}-s_{n}|\\ &\leq|s_{n+p}-s_{n+p-1}|+\cdots+|s_{n+1}-s_{n}|\\ &<\dfrac{1}{2^{n+p-1}}+\cdots+\dfrac{1}{2^{n}}. \end{align*} Note that $\displaystyle\sum\dfrac{1}{2^{n}}<\infty$.

Answer 3

$$|s_{k+1}-s_k|=|a_{k+1}|<\frac1{2^{k+1}}$$

Hence by comparison test, it converges absolutely.