I have troubles in finding a formula for the partial sums of this series:

$\sum_{k=1}^\infty \frac{k}{2^k} = 2 $

How do I have to proceed?



$$\sum_{i=0}^n x^i = \frac{x^{n+1}-1}{x-1}$$

Now differentiate both members with respect to $x$, multiply both members by $x$ and you get the general formula. For your case just set $x = \frac 12$

  • $\begingroup$ @user21820 thanks! edited $\endgroup$ – Ant Mar 24 '15 at 8:29

Here is one way to do it without differentiation:

$\sum_{k=1}^\infty k f(k)$

$= \sum_{k=1}^\infty \sum_{m=1}^k f(k)$

$= \sum_{1 \le m \le k \le \infty} f(k)$

$= \sum_{m=1}^\infty \sum_{k=m}^\infty f(k)$

Substitute the appropriate function $f$ and I'm sure you can now evaluate the inner sum and then the outer sum.

  • $\begingroup$ Any idea if this is somewhat related to expected value? I think this was in a probability book I saw $\endgroup$ – user198044 Mar 24 '15 at 8:47
  • 1
    $\begingroup$ @Jack: Indeed! This technique finds itself in probability as well! I can't recall what it is called but something like tail expectation.. $\endgroup$ – user21820 Mar 24 '15 at 8:49
  • $\begingroup$ 21820, I think this is expected value of a random variable whose range is $\mathbb{N}$ ? $\endgroup$ – user198044 Mar 24 '15 at 8:51
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    $\begingroup$ @Jack: I guess you can say so, for discrete natural valued random variable. The same technique applies to the continuous case as well. $\endgroup$ – user21820 Mar 24 '15 at 8:54

Observe that your series can be rewritten as

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

Whenever you see something like a variable multiplied by something raised to that variable, consider differentiating some other series. In our case, let's use:

$\frac{d}{dx} \sum_{n=0}^{\infty} (x)^n = \sum_{n=1}^{\infty} nx^{n-1}$

You may recall some formula that looks like

$\sum_{n=0}^{\infty} (x)^n = \frac{1}{1-x}$

Unfortunately, this is not true $\forall x \in \mathbb{R}$.

It is true provided $-1 < x < 1$. In our case, we can have x = 1/2 and then use the formula.

Thus, we have

$\sum_{n=0}^{\infty} (x)^n = \frac{1}{1-x}$

$\frac{d}{dx} \sum_{n=0}^{\infty} (x)^n = \frac{d}{dx} \frac{1}{1-x}$

$\sum_{n=0}^{\infty} nx^{n-1} = \frac{1}{(1-x)^{2}}$

Don't plug in x = 1/2 at this point.

Make the LHS the same as $\sum_{k=0}^{\infty} k (1/2)^k = \sum_{k=0}^{\infty} k (x)^k|_{x=1/2}$ by multiplying x on both sides:

$x \sum_{n=0}^{\infty} nx^{n-1} = x \frac{1}{(1-x)^{2}}$

$\sum_{n=0}^{\infty} nx^{n} = \frac{x}{(1-x)^{2}}$

Not done. Note the index. It should start at 1 not 0.

$0 + \sum_{n=1}^{\infty} nx^{n} = \frac{x}{(1-x)^{2}}$

Now plug in x = 1/2.

Now if you want to compute $\sum_{k=1}^{n} k (1/2)^k$, the process is similar.

$\sum_{k=1}^{n} (x)^k = \frac{1-x^{n+1}}{1-x}$

We differentiate to get:

$\sum_{k=1}^{n} k (x)^{k-1} = \frac{d}{dx} \frac{1-x^{n+1}}{1-x}$

Now that there's an x in the numerator I don't feel like doing this hahaha. Hope you get the idea


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