1. ## convergence question

$\displaystyle \sum _{h=0}^\infty \frac{h}{2^h}$
by what method it converges?
to what it converges?

2. ## Re: convergence question

Originally Posted by transgalactic
$\displaystyle \sum _{h=0}^\infty \frac{h}{2^h}$
by what method it converges?
to what it converges?
Hi transgalactic!

The standard method I know is to define $\displaystyle f(x) = \sum_{h=0}^\infty {h x^h}$.
Then your series corresponds to f(1/2).

If you integrate f(x) you should be able to find its limit (using the formula for a geometric series).
If you differentiate the result, you find f(x).

Alternatively, you can simply write out a number of terms.
Rearrange the terms in a triangle.
Find the sums of each row in the triangle.
And sum the results.

3. ## Re: convergence question

hi transgalactic

here are some tests for determining whether series are convergent or not: Convergence Tests -- from Wolfram MathWorld

EDIT: if it were me i would either use the ratio test (Ratio Test -- from Wolfram MathWorld)

4. ## Re: convergence question

Root test is faster.

5. ## Re: convergence question

i did the a_n+1/a_n test
and got 1/2

so it converges to 1/2
?

6. ## Re: convergence question

Converges but that's not the value.

7. ## Re: convergence question

to what it converges?

8. ## Re: convergence question

Originally Posted by transgalactic
to what it converges?
$\displaystyle \begin{gathered} \sum\limits_{k = 0}^\infty {x^k } = \frac{1}{{\left( {1 - x} \right)}},\;|x| < 1 \hfill \\ \sum\limits_{k = 0}^\infty {kx^{k - 1} } = \frac{1}{{\left( {1 - x} \right)^2 }} \hfill \\ \sum\limits_{k = 0}^\infty {kx^k } = \frac{x}{{\left( {1 - x} \right)^2 }} \hfill \\ \end{gathered}$

9. ## Re: convergence question

Originally Posted by transgalactic
to what it converges?
Another way: in general, $\displaystyle \sum_{n=1}^{+\infty}nr^n$ converges if $\displaystyle |r|<1$ (use for example the ratio test). Denote $\displaystyle S$ the sum of the series, we have:

$\displaystyle (i)\;\;S=r+2r^2+3r^3+\ldots \quad (ii)\;\; rS=r^2+2r^3+3r^4+\ldots$

This implies $\displaystyle S-rS=r+r^2+r^3+\ldots=\frac{1}{1-r}-1=\frac{r}{1-r}$ as a consequence $\displaystyle S=\frac{r}{(1-r)^2}$ . For your case, substitute $\displaystyle r=1/2$ .