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.
Best 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 $$H_k=\sum_{n=1}^k\frac1n$$