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 $\sum_{k = 1}^n\dfrac{1}{k}(1/2)^{n - k}$.

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 $\displaystyle{S_n \sim \frac{2}{n}}$.
