How to show $\sum\limits_{i=1}^{t}\frac{1}{i}2^{t-i}=2^t\ln 2 -\frac{1}{2}\sum\limits_{k=0}^\infty \frac{1}{2^k(k+t+1)}$

Question

How to show the below equation ?

$$\sum\limits_{i=1}^{t}\frac{1}{i}2^{t-i}=2^t\ln 2 -\frac{1}{2}\sum\limits_{k=0}^\infty \frac{1}{2^k(k+t+1)} ~~~~~(t\in \mathbb Z^+)$$

Answer 1

One may write

$$ \begin{align} \sum\limits_{k=0}^\infty \frac{1}{2^k(k+t+1)}&=\sum\limits_{i=t+1}^\infty \frac{1}{2^{i-t-1}i}\qquad (i=k+t+1)\\\\ &=\sum\limits_{i=1}^\infty \frac{1}{2^{i-t-1}i}-\sum\limits_{i=1}^t \frac{1}{2^{i-t-1}i}\\\\ &=2^{t+1}\sum\limits_{i=1}^\infty \frac{1}{2^{i}i}-2^{t+1}\sum\limits_{i=1}^t \frac{1}{2^ii}\\\\ &=2^{t+1}\ln 2-2^{t+1}\sum\limits_{i=1}^t \frac{1}{2^ii}, \end{align} $$ then divide by two, where we have used the classic Taylor expansion $$ -\ln (1-x)=\sum\limits_{i=1}^\infty \frac{x^i}{i},\quad |x|<1. $$