Series convergence

Max.Planck

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).

[itex]\sum_{i=1}^\infty \frac{i}{2^i}[/itex]

micromass

Do you know the sum of

[tex]\sum_{n=1}^{+\infty} x^n[/tex]

for |x|<1 ??

Now take derivatives.

Max.Planck

I know the sum, it evaluates to:
[tex]\frac{1}{1-x}-1[/tex]

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

[tex]\sum_{n=1}^{\infty} \frac{n!}{(n-i)!}x^{n-i}[/tex]

DonAntonio

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:
[tex]\frac{1}{1-x}-1=\sum_{n=1}^\infty x^n\Longrightarrow \frac{1}{(1-x)^2}=\sum_{n=1}^\infty nx^{n-1}[/tex]

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

DonAntonio

Millennial

I offer another way (not very rigorous, but still intuitive.)
The nth term of the sum can be written as
[tex]\left(\frac{1}{2^n}+\frac{1}{2^n}...+\frac{1}{2^n}\right)[/tex]
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!

Max.Planck

But how do you choose that x? Shouldn't

[tex]x^{n-1}=(1/2)^{n}[/tex]

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

DonAntonio

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

DonAntonio

Max.Planck

Ah right lol, thank you very much.