What is the limit of this series? assume  $$x_n=\frac{n+1}{2^{n+1}}\sum_{k=1}^n\frac{2^k}{k} ,n=1,2,.....$$
how compute $\lim_{n\to +\infty}x_n$?
Thanks for any hint
 A: Since the answer is straightforward if we exploit the Cesaro-Stolz theorem, here I want to present a solution that does not use this theorem.
Notice first that whenever $1<r<2$, we have
$$ \lim_{n\to\infty} \frac{\sum_{k＝1}^{n} \frac{r^k}{k}}{\frac{2^{n+1}}{n+1}} ＝ 0.$$
Indeed, the numerator is dominated by $n r^n$, thus we have
$$ 0 \leq \frac{\sum_{k＝1}^{n} \frac{r^k}{k}}{\frac{2^{n+1}}{n+1}} \leq \frac{n(n+1)}{2} \left(\frac{r}{2}\right)^n \xrightarrow[]{n\to\infty} 0.$$
This shows that, the limit in question exists with value $\ell$ if and only if
$$ \lim_{n\to\infty} \frac{\sum_{k＝1}^{n} \frac{2^k - r^k}{k}}{\frac{2^{n+1}}{n+1}} ＝ \ell.$$
But note that
$$ \sum_{k＝1}^{n} \frac{2^k - r^k}{k} ＝ \sum_{k＝1}^{n} \int_r^2 x^{k-1} \; dx ＝ \int_r^2 \frac{x^{n} - 1}{x-1} \; dx. $$
This in particular shows that
$$ \sum_{k＝1}^{n} \frac{2^k - r^k}{k} \leq \int_r^2 \frac{x^{n}-1}{r-1} \; dx \leq \frac{2^{n+1}-r^{n+1}-(n+1)(2-r)}{(r-1)(n+1)} $$
and
$$ \sum_{k＝1}^{n} \frac{2^k - r^k}{k} \geq \int_r^2 (x^{n}-1) \; dx \geq \frac{2^{n+1}-r^{n+1}-(n+1)(2-r)}{n+1} $$
These equalities show that
$$ 1\leq \liminf_{n\to\infty} \frac{\sum_{k＝1}^{n} \frac{2^k - r^k}{k}}{\frac{2^{n+1}}{n+1}} \leq \limsup_{n\to\infty} \frac{\sum_{k＝1}^{n} \frac{2^k - r^k}{k}}{\frac{2^{n+1}}{n+1}} \leq \frac{1}{r-1}. $$
Now since $r$ was arbitrary, taking $r\to2^{-}$ yields the limit $\ell ＝ 1$.
