The sum $\sum_{k=0}^\infty (k-1)/2^k$ is zero

Question

I am trying to prove an equation given in the CLRS exercise book (AP GP clrs appendix A.1-4). The equation is:

$$\sum_{k=0}^\infty (k-1)/2^k=0$$

I solved the LHS but my answer is 1 whereas the RHS should be 0 Following is my solution:

Let's say S = k/2^k = 1/2 + 2/2^2 + 3/2^3 + 4/2^4 ....
2S = 1 + 2/2 + 3/2^2 + 4/2^3 ...

2S - S = 1 + ( 2/2 - 1/2) + (3/2^2 - 2/2^2) + (4/2^3 -     3/2^3)..
S = 1+ 1/2 + 1/2^2 + 1/2^3 + 1/2^4..
S = 2       -- eq 1

Now let's say S1 = (k-1)/2^k = 0/2 + 1/2^2 + 2/2^3 + 3/2^4...

S - S1 = 1/2 + (2/2^2 - 1/2^2) + (3/2^3 - 2/2^3) + (4/2^4 - 3/2^4)....
S - S1 = 1/2 + 1/2^2 + 1/2^3 + 1/2^4...
       = 1
From eq 1
2 - S1 = 1
S1 = 1

Whereas the required RHS is 0. Is there anything wrong with my solution? Thanks..

You showed that $\sum\limits_{k=1}^\infty \frac{k-1}{2^k} = 1$, yet the question was about the sum starting from $k=0$ instead of $k=1$. — JMoravitz, Jun 14 '15 at 18:03

Error 404 · Accepted Answer · 2015-06-15T02:18:37.527

1

You forgot to add the term $-1$ in $S_1$. It is achieved at $k=0$. So, $ S-S_1 = 2.$

edited Jun 15 '15 at 02:18

answered Jun 14 '15 at 16:28

Error 404

6,006

Got it.. Thanks a lot :) – Saurabh Jun 14 '15 at 18:48

The sum $\sum_{k=0}^\infty (k-1)/2^k$ is zero

1 Answers1

Linked