Proving the Convergence of ∑ n/2^n to 2

[Lamarkiz]

I know that the following serie converges to 2 (did in excel), still I would like to know how i can prove it step by step it.

∞
∑ n/2^n
n=1

I tried (n+1)/(2^(n+1))/(n/2^n) still I'm finding 1/2, not the 2.

Any thoughts?

 

[micromass]

Well, do you know what

\sum_{n=0}^{+\infty}{\frac{x^n}{2^n}}

converges to??

And what if you take the derivative of that??

 

[Lamarkiz]

Well, do you know what

\sum_{n=0}^{+\infty}{\frac{x^n}{2^n}}

converges to??

And what if you take the derivative of that??

Sorry, can you be more specific please?

 

[micromass]

Sorry, can you be more specific please?

The series

\sum_{n=0}^{+\infty}{\frac{x^n}{2^n}}

Can you find out what it converges to?? It's a nice geometric series...

 

[Lamarkiz]

The series

\sum_{n=0}^{+\infty}{\frac{x^n}{2^n}}

Can you find out what it converges to?? It's a nice geometric series...

I did them in excel, found the following values.

\sum_{n=0}^{+\infty}{\frac{n}{2^n}} = 2/1

\sum_{n=0}^{+\infty}{\frac{n}{3^n}} = 3/4

\sum_{n=0}^{+\infty}{\frac{n}{4^n}} = 4/9

\sum_{n=0}^{+\infty}{\frac{n}{5^n}} = 5/16

and on...

So i figured out that i can find where any of these series (n/x^n) converge doing: x/(x-1)² where |x|<1 (it works to negative values too)

I showed it to my professor, he says it's only a form of approximation, that I'm not proofing anything at all.

The funny fact is that i can calculate

\sum_{n=0}^{+\infty}{\frac{n}{-1024^n}}

And i know the value that serie will converge will be given by : (-1024/(-1024-1)²) = -9.74657941701368E-4

Still i need a way to get to this... or a simple resolution of the previous given serie will help too.

 

[micromass]

Read this first:

http://en.wikipedia.org/wiki/Geometric_series

Can you then find the sum of

\sum_{n=0}^{+\infty}{\frac{x_n}{2^n}}

 

[Lamarkiz]

Read this first:

http://en.wikipedia.org/wiki/Geometric_series

Can you then find the sum of

\sum_{n=0}^{+\infty}{\frac{x_n}{2^n}}

Sorry, spent like 1 hour in it and still can't find it. I'm tired today, thanks for the help anyway. I'm going to try tomorrow.

 

[HallsofIvy]

Generally, people learn, in, say, a PreCalculus class, if not a basic algebra class, long before Calculus, that, if |r|< 1, then the geometric series
\sum_{n= 0}^\infty ar^n= \frac{a}{1- r}

That is what micromass has been trying to remind you of. The series
\sum \left(\frac{x}{2}\right)^n
is a geometric sequence with a= 1, r= x/2. As long as |x/2|< 1, i.e., as long as -2< x< 2, you can find the sum by the formula above.

And, since
\frac{d x^n}{dx}= nx^{n-1}
the derivative of that function has series
\sum \frac{nx^n}{2^n}
which, taking x= 1, becomes
\sum \frac{n}{2^n}