sequences and series - Infinite sum of harmonic number

I learned that I can find the value of some infinite sum.

Then what is the value of this sum?
$$\frac12 + \left(1+\frac12\right)\frac1{2^2}+\left(1+\frac12 +\frac13\right)\frac1{2^3}+\left(1+\frac12 +\frac13 +\frac14\right)\frac1{2^4} + \cdots $$
And I want to know How to find the value of the infinite sum of this-like form.

Answer

One can split this summation into
$$\left(\frac12+\frac1{2^2}+\frac1{2^3}+\cdots\right)+\frac12\left(\frac1{2^2}+\frac1{2^3}+\cdots\right)+\frac13\left(\frac1{2^3}+\frac1{2^4}+\cdots\right)+\cdots$$
$$\begin{align}
&=\sum_{k=1}^\infty \frac1k \sum_{j=k}^\infty \frac1{2^j}\\
&=\sum_{k=1}^\infty \frac1k \left(\frac{2^{-k}}{1-2^{-1}}\right)\\
&=\sum_{k=1}^\infty \frac{2^{1-k}}k\\
&=2\sum_{k=1}^\infty \frac{\left(\frac12\right)^k}k\\
&=2\left(-\ln{\left(1-\frac12\right)}\right)\\
&\boxed{=2\ln{(2)}}

\end{align}$$
By using the fact that
$$\ln{(1-x)}=-\sum_{k=1}^\infty \frac{x^k}k$$
for all $|x|\lt 1$.

In fact one can use a similar method to prove that
$$\sum_{k=1}^\infty x^kH_k=\frac1{1-x}\ln{\left(\frac1{1-x}\right)}$$
for $|x|\lt1$. Where $H_k$ is the $k$th harmonic number given by
$$H_k=\sum_{n=1}^k\frac1n$$