Is the sequence with the property $|a_{n+1} - a_n| \le \frac{n}{2^n}$ Cauchy?

Question

I was looking at this other question and was wondering if the following holds.

The sequence that has the property that each term $\{a_n\}$ satisfies $|a_{n+1} - a_n| \le \frac{n}{2^n}$ for all $n$, is Cauchy.

My initial thought was no. Wlog let $n>m$,and we would get something like (after simplifying in an analogous manner to the accepted answer in the above mentioned post), $$|a_n-a_m|<\frac{n}{2^{m-1}}$$ and you're pretty much stuck at this point, I think. But I'm not able to definitively say it is not Cauchy.

Any advice on wether we can save this argument and prove the sequence is Cauchy? Or maybe it isn't?

Answer 1

A possible standard trick for such questions is to note that

$$a_n = a_0 + \sum_{k=1}^n(a_k-a_{k-1})$$

Now, if $\sum_{k=1}^n(a_k-a_{k-1})$ is convergent, then $a_n$ must be convergent, as well, and is Cauchy.

Since $$|\sum_{k=1}^n(a_k-a_{k-1})| \leq \sum_{k=1}^n|a_k-a_{k-1}|\leq \sum_{k=1}^n\frac k{2^k}$$

the series is (absolutely) convergent.

Answer 2

i) $|a_n -a_{n+1}| < r^n,\ r<1$ is Cauchy :

Given $\varepsilon$, there is $N$ i.e. $r^N < \varepsilon $ s.t. $n> m>N$ implies $$ |a_n -a_m| < |a_n -a_{n-1} | +\cdots +| a_{m+1} -a_m | < r^{n-1} +\cdots +r^m < r^N \frac{1}{1-r} < \frac{ \varepsilon }{1-r}$$ ii) Here $\frac{n}{\sqrt{2}^n} <1 ,\ \forall n >N$ for some $N$.

Since $\frac{1}{\sqrt{2}}<1$, then given sequence in OP is Cauchy.