Finding the sum of a serie by recognizing it as a Mclaurin serie

[tamtam402]

Homework Statement

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

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

Homework Equations

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

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!

 

[Mark44]

Homework Statement

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

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

Homework Equations

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

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:
$$\sum_{n = 1}^{\infty} \frac{1}{2^n} $$

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!

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.

 

[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.