# Entropy of geometric random variable with parameter $1/2$

I'm studing for an information theory exam, maybe some of you can help me here with an exercise.

What's the entropy of $X$ as $\{1,2,\ldots,n\}$ ($n$=infinity) where the probabilities are $P \{1/2^1, 1/2^2,\ldots, 1/2^n\}$?

The question is multiple choice and gives 4 possible answers: 1. $2 \over 3$ bits/symbol; 2. $1 \over 2$ bits/symbol; 3. $\infty$ bits/symbol; 4. none of the above;

So far i got: $$H(X) = - \sum_{i=1}^{n} P(x_i) \cdot\log_2( P(x_i))$$

So in this case, $$H(X) = - \sum_{i=1}^{\infty} {1 \over 2^i} \cdot\log_2\left({1 \over 2^i}\right)$$

$\log_2(1/x) = -\log_2(x)$, while $x>0$, so,

$$H(X) = - \sum_{i=1}^{\infty} {1 \over 2^i}\cdot(-i)$$

I also know that:

$$\sum_{i=1}^{\infty} a \cdot r^{-i} = {a \over r-1}$$

But in this case I think 'a' must be a constant, right?

Wolfram Alpha gives me H(X) = 2 bits/symbol as the result: bit.ly/nbQwgV

It is correct? Any hint?

Greatly apreciated. Cheers.

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In the last series $i$ starts at $0$. It should be $\displaystyle\sum_{i=0}^{\infty}a\cdot r^i=\dfrac{a}{1-r}$. –  Américo Tavares Jul 10 '11 at 17:35
The first two options can be eliminated by adding up a few terms. To eliminate infinity, one needs some intuition, that the $1/2^i$ go down very fast, enough to neutralize the $i$ on top. So knowing a closed form for $\sum \frac{i}{2^i}$ is not necessary to decide on the answer. But the closed form is nice. –  André Nicolas Jul 10 '11 at 23:49

You need to know the value of the series $t(x)=\displaystyle\sum_{i\ge0}ix^i$. You probably already know the value of the series $s(x)=\displaystyle\sum_{i\ge0}x^i$ since $s(x)$ is the classical geometric series: $s(x)=1/(1-x)$.
But $t(x)=xs'(x)$ hence $t(x)=x/(1-x)^2$.
In your case, you start at $i=1$ instead of $i=0$ but this does not change anything since the $0$th term of $t(x)$ is $0$, and you choose $x=\frac12$. Hence $t(\frac12)=\frac12/(1-\frac12)^2=2$, as desired.