<s>[INST] 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 $\sum_{k = 1}^n\dfrac{1}{k}(1/2)^{n - k}$.
 [/INST] \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.
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