# Homework Help: Finding the sum of a serie by recognizing it as a Mclaurin serie

1. May 21, 2012

### tamtam402

1. The problem statement, all variables and given/known data

I'm trying to find the sum of the following serie:

$\Sigma^{\infty}_{0}\frac{1}{n2^{n}}$

2. Relevant equations

$ln(1+x)=\Sigma^{\infty}_{0}\frac{(-1)^{n+1}x^{n}}{n}$

3. The attempt at a solution

According to my book, the serie is equal to ln(2), which means I should be able to substitute x = 1 in the serie development of ln(1+x) and find the form of the serie. However, I can't see how the two are similar. There isn't even a (-1)^n in the serie I'm asked to find the sum of!

2. May 21, 2012

### Staff: Mentor

Both series are undefined if n = 0, so I would guess that they both need to start at 1.

By the way, "series" is both singular a plural in English - there is no word "serie" in English.

In LaTeX, you can also use \sum for a series, like this:
[ tex ] \sum_{n = 1}^{\infty} \frac{1}{2^n}[ /tex ]
If the extra spaces are removed, it renders as this:
$$\sum_{n = 1}^{\infty} \frac{1}{2^n}$$
Maybe there's a typo in your book. It looks to me like x = -1/2 would get rid of all but one factor of -1.

3. May 21, 2012

### Bohrok

If x = -1/2, then ln(1 - 1/2) = ln(1/2) = ln(2-1) = -ln(2)
Then by putting -1/2 for x in the series expansion of ln(1+x), xn will alternate in sign and you can rewrite the sum with just positive terms.