Is the series $\sum_{n=1}^\infty\frac{1}{n}\left(\sum_{k=1}^n\frac{1}{k}\left(\frac{1}{2}\right)^{n-k}\right)$ convergent? $$
\sum_{n = 1}^\infty\dfrac{1}{n}\left(\sum_{k = 1}^n\dfrac{1}{k}\left(\dfrac{1}{2}\right)^{n - k}\right)
$$
Does the series converge?
I calculate it using Matlab, and it seems that the sum converges to 2.4673. I also tried to use the ratio test to prove the convergent, but stopped at to calculate the sum $\sum_{k = 1}^n\dfrac{1}{k}(1/2)^{n - k}$.
 A: Just to add to the posted nice solutions: the exact value for the sum can be obtained. Knowing already that the series converges, and changing the order of summation and integration, we can get the answer.
If we denote $f(a,n)=\sum_{k=1}^n\frac{e^{ak}}{k}\,\,\Rightarrow\,\,f'(x,n)=\frac{\partial}{\partial x}f(x,n)=\sum_{k=1}^n e^{xk}$ and $f(a,n)=\int_{-\infty}^a f'(x,n)dx$.
Then the initial sum
$$S=\sum_{n = 1}^\infty\dfrac{1}{n}\left(\sum_{k = 1}^n\dfrac{1}{k}\left(\dfrac{1}{2}\right)^{n - k}\right)=\sum_{n=1}^\infty\frac{1}{n}\frac{1}{2^n}\sum_{k=1}^n\frac{2^k}{k}=\sum_{n=1}^\infty\frac{1}{n}\frac{1}{2^n}f(\ln2,n)$$
$$=\sum_{n=1}^\infty\frac{1}{n}\frac{1}{2^n}\int_{-\infty}^{\ln2}f'(x,n)dx$$
Evaluating $f'(x,n)$
$$f'(x,n)=\sum_{k=1}^n e^{xk}=\frac{e^x(1-e^{xn})}{1-e^x}$$
Therefore, our sum can be presented in the form
$$S=\sum_{n=1}^\infty\frac{1}{n}\frac{1}{2^n}\int_{-\infty}^{\ln2}\frac{e^x(1-e^{xn})}{1-e^x}dx=\int_{-\infty}^{\ln2}\frac{e^x}{1-e^x}dx\bigg(\sum_{n=1}^\infty\frac{1}{n}\frac{1}{2^n}-\sum_{n=1}^\infty\frac{1}{n}\frac{1}{2^n}e^{nx}\bigg)$$
$$=\int_{-\infty}^{\ln2}\frac{e^x}{1-e^x}dx\bigg(-\ln\Big(1-\frac{1}{2}\Big)+\ln\Big(1-\frac{e^x}{2}\Big)\bigg)=\int_{-\infty}^{\ln2}\frac{e^x}{1-e^x}\ln\big(2-e^x\big)dx$$
Making the substitution $e^x=t$
$$S=\int_0^2\frac{\ln(2-t)}{1-t}dt$$
Making another substitution $x=1-t$
$$S=\int_{-1}^1\frac{\ln(1+x)}{x}dx=2\sum_{k=0}^\infty\frac{1}{(2k+1)^2}$$
Given that $\sum_{k=0}^\infty\frac{1}{(2k+1)^n}=\zeta(n)(1-2^{-n})$, we get the answer:
$$\boxed{S=2\zeta(2)(1-2^{-2})=\frac{\pi^2}{4}}\,\,(=2.467...)$$
A: By the Stolz-Cesàro Theorem, we have $$\sum_{k=1}^n \frac{2^k}{k} \sim \frac{2^{n+1}}{n}$$
So $$\frac{1}{n}\sum_{k=1}^n\frac{1}{k}\left(\frac{1}{2}\right)^{n-k} = \frac{1}{n\, 2^n}\sum_{k=1}^n \frac{2^k}{k} \sim \frac{1}{n\,2^n}\frac{2^{n+1}}{n} = \frac{2}{n^2}$$
Hence, the series converge.
A: Let
$$S_n = \sum_{k=1}^n \frac{1}{k} \left( \frac{1}{2} \right)^{n-k}$$
Let's show that $(nS_n)$ is eventually decreasing. By direct computation, for $n > 2$
\begin{align*} (n-1)S_{n-1} – nS_n &= \left( \sum_{k=1}^{n-1} \frac{n-1}{k} \left( \frac{1}{2} \right)^{n-1-k}\right)- \left(\sum_{k=1}^n \frac{n}{k} \left( \frac{1}{2} \right)^{n-k} \right)\\
&= \left( \sum_{k=1}^{n-1} \frac{n-1}{k} \left( \frac{1}{2} \right)^{n-1-k}\right)- \left(\sum_{k=0}^{n-1} \frac{n}{k+1} \left( \frac{1}{2} \right)^{n-k-1} \right)\\
&=\left[ \sum_{k=1}^{n-1} \left(\frac{n-1}{k} - \frac{n}{k+1}\right)\left( \frac{1}{2} \right)^{n-1-k}\right]- \frac{n}{2^{n-1}}\\
&=\left( \sum_{k=1}^{n-1} \frac{n-k-1}{k(k+1)2^{n-1-k}} \right)- \frac{n}{2^{n-1}} \\
&=\left( \sum_{k=0}^{n-2} \frac{k}{2^k(n-k)(n-k-1)} \right) - \frac{n}{2^{n-1}}\quad \quad (\text{substitution } k \rightarrow n-1-k)\\
 &= \left( \sum_{k=2}^{n-2} \frac{k}{2^k(n-k)(n-k-1)} \right)  + \left(\frac{1}{2(n-1)(n-2)}- \frac{n}{2^{n-1}} \right)
\end{align*}
The second parenthesis is clearly positive for $n$ sufficiently large, hence $(n-1)S_{n-1} – nS_n > 0$ for $n$ sufficiently large. In particular, the sequence $(nS_n)$ is bounded : there exists $M \geq 0$ such that $nS_n \leq M$. In particular, one has
$$0 \leq \frac{1}{n} S_n \leq \frac{M}{n^2}$$
By comparison, you get that the series $\displaystyle{\sum\frac{1}{n} S_n}$ converges.

Remark : Actually, one can prove using the recursion
$$S_n = \frac{1}{2}S_{n-1} + \frac{1}{n}$$
that the sequence $(nS_n)$ converges to $2$, so $\displaystyle{S_n \sim \frac{2}{n}}$.
A: \begin{aligned}\sum_{k=1}^{n}{\frac{2^{k-n}}{k}}&=\frac{1}{2^{n-1}}\sum_{k=1}^{n}{\int_{0}^{1}{\left(2x\right)^{k-1}\,\mathrm{d}x}}\\ &=\frac{1}{2^{n-1}}\int_{0}^{1}{\frac{1-2^{n}x^{n}}{1-2x}\,\mathrm{d}x}\\ &=\frac{1}{2^{n}}\int_{-\frac{1}{2}}^{\frac{1}{2}}{\frac{1-\left(1-2x\right)^{n}}{x}\,\mathrm{d}x}\end{aligned}
The integral in the second line make sense because $ x\mapsto\frac{1-\left(2x\right)^{n}}{1-2x} $ can be extended by continuity at $ \frac{1}{2} $, which means it is piecewise continuous on $ \left[0,1\right]$, and, hence, is integrable.
In the third line We've just used a substitution $ y=\frac{1}{2}-x $. Now using the fact that for $ \alpha\geq 0 $, we have that $ \left(\forall x\geq -1\right), \left(1+x\right)^{\alpha}\geq 1+\alpha x $. (Bernoulli's inequality), we have : \begin{aligned}\frac{1}{2^{n}}\int_{-\frac{1}{2}}^{\frac{1}{2}}{\frac{1-\left(1-2x\right)^{n}}{x}\,\mathrm{d}x}\leq\frac{1}{2^{n}}\int_{-\frac{1}{2}}^{\frac{1}{2}}{\frac{2nx}{x}\,\mathrm{d}x}=\frac{n}{2^{n-1}}\end{aligned}
Since $ \sum\limits_{n\geq 1}{\frac{1}{2^{n-1}}} $ converges, our series also converges.
