I am trying to prove an equation given in the CLRS exercise book. The equation is:

```
Sigma k=0 to k=infinity (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..

Answer:

Yes, you have issues in your solution to the problem.

While everything is correct in formulating the value of S, you have calculated the value of S1 incorrectly. You missed substituting the value for k=0 in S1. Whereas, for S, even after putting the value of k, the first term will be 0, so no effect.

Therefore,

```
S1 = (k-1)/2^k = -1 + 0/2 + 1/2^2 + 2/2^3 + 3/2^4...
// you missed -1 here because you started substituting values from k=1
S - S1 = -(-1) + 1/2 + (2/2^2 - 1/2^2) + (3/2^3 - 2/2^3) + (4/2^4 - 3/2^4)....
S - S1 = 1 + (1/2 + 1/2^2 + 1/2^3 + 1/2^4...)
= 1 + 1
= 2.
From eq 1
2 - S1 = 2
S1 = 0.
```

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