# Series convergence

Hi, while reading some artificial intelligence book, i came upon the following sum. How can I evaluate it analytically, so not guess it by computing many terms? It's easy to see by ratio test that it converges (intuitively too, since its a linear vs exponential function).

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

Do you know the sum of

$$\sum_{n=1}^{+\infty} x^n$$

for |x|<1 ??

Now take derivatives.

Do you know the sum of

$$\sum_{n=1}^{+\infty} x^n$$

for |x|<1 ??

Now take derivatives.

I know the sum, it evaluates to:
$$\frac{1}{1-x}-1$$

But, how will the derivatives help? The i-th derivative with respect to x gives me:

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

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I know the sum, it evaluates to:
$$\frac{1}{1-x}-1$$

But, how will the derivatives help? The i-th derivative with respect to x gives me:

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

Why would you take the i-th derivative if you were said "the derivative? I find that i as index pretty confussing and annoying, so I'll change it:
$$\frac{1}{1-x}-1=\sum_{n=1}^\infty x^n\Longrightarrow \frac{1}{(1-x)^2}=\sum_{n=1}^\infty nx^{n-1}$$

and now just choose a convenient x within the convergence radius...

DonAntonio

I offer another way (not very rigorous, but still intuitive.)
The nth term of the sum can be written as
$$\left(\frac{1}{2^n}+\frac{1}{2^n}...+\frac{1}{2^n}\right)$$
where we have the fraction repeated n times. Now look at this structure:

1/2
1/4+1/4
1/8+1/8+1/8
.............

Now sum columns instead of the rows!

Why would you take the i-th derivative if you were said "the derivative? I find that i as index pretty confussing and annoying, so I'll change it:
$$\frac{1}{1-x}-1=\sum_{n=1}^\infty x^n\Longrightarrow \frac{1}{(1-x)^2}=\sum_{n=1}^\infty nx^{n-1}$$

and now just choose a convenient x within the convergence radius...

DonAntonio

But how do you choose that x? Shouldn't

$$x^{n-1}=(1/2)^{n}$$

But then x will be a term dependent on n...

But how do you choose that x? Shouldn't

$$x^{n-1}=(1/2)^{n}$$

But then x will be a term dependent on n...

...and thus $\,\displaystyle{ x^{n-1}=\frac{1}{2}\left(\frac{1}{2}\right)^{n-1}}$...! And of course, taking off a factor of 0.5, x is a constant.

DonAntonio

...and thus $\,\displaystyle{ x^{n-1}=\frac{1}{2}\left(\frac{1}{2}\right)^{n-1}}$...! And of course, taking off a factor of 0.5, x is a constant.

DonAntonio

Ah right lol, thank you very much.