I am a computer scientist trying resolve exercises from CLRS. Here is one that I can't make progress on.
Show that $\sum\limits_{k=0}^{\infty} (k-1)/2^k = 0 $
What I did so far:
$$ \sum_{k=0}^{\infty} \frac{k}{2^k} – \sum_{k=0}^{\infty} \frac{1}{2^k} $$
And by this, seems that is not 0 the correct answer.
(This is appendix question A.1-4 from Introduction to Algorithms by Cormen 3ed)
Best Answer
The answer is $0$. To see this. Let's recall the identities
$$ \sum_{k=0}^{\infty} x^k = \frac{1}{1-x},\quad \sum_{k=0}^{\infty} k x^k = \frac{x}{(1-x)^2} $$
which gives
$$ \sum_{k=0}^{\infty} k x^k -\sum_{k=0}^{\infty} x^k = \frac{2x-1}{(1-x)^2} =0 \iff 2x-1=0 \iff x = \frac{1}{2}$$