If $x_{n^2}= 2x_{2n}+n^{1/n} - \sum\limits_{k=1}^{\infty}\frac{1}{2^k}$ and $\sum(x_{n+1}-x_n)$ converges, then $\lim x_n = 0$ Let $x_n$ be a real number sequence such as: $x_{n^2}= 2x_{2n}+n^{1/n} - \sum_{k=1}^{\infty}\frac{1}{2^k}$, for every $n=1,2,3...$ and the sum $\sum_{n=1}^{\infty}(x_{n+1}-x_n)$ converges. Show that $\lim x_n = 0$.
So, I began by saying that since the sum converges then, $(x_{n+1}-x_n) \to0$ and then I tried to find the $x_{(n+1)^2}-x_{n^2}$ difference and then take its limit saying that it must be equal to 0 (Not sure about that step). With that, I have:  $\lim (2x_{2(n+1)}-2x_{2n}+(n+1)^{1/n+1} - n^{1/n}) = 0 \implies \lim (x_{2(n+1)} -x_{2n}) = 0 $ and i don't know what to do from here.

Note that:


*

*The series $\sum(x_{n+1}-x_n)$ converges and its $n$th partial sum is $x_{n+1}-x_1$, hence the sequence $(x_n)$ converges, call $\ell$ its limit.

*In the identity $x_{n^2}= 2x_{2n}+n^{1/n} - \sum\limits_{k=1}^{\infty}\frac{1}{2^k}$, the term $\sum\frac1{2^k}$ equals $1$ and $n^{1/n}\to1$ when $n\to\infty$.

*Since $x_{n^2}\to\ell$ and $x_{2n}\to\ell$, item 2. implies that $\ell=2\ell+1-1$.

*Because $\ell$ is finite, item 3. implies that $\ell=0$.


QED.
