# prove convergence and find sum

• Jul 5th 2008, 02:15 PM
galactus
prove convergence and find sum
Let $a_{n}=1-\frac{1}{2}+\frac{1}{3}-....+\frac{(-1)^{n-1}}{n}-ln(2)$

Prove $\sum_{n=1}^{\infty}a_{n}$ converges and find its sum.

This shouldn't be too bad. Just a neat series.

I started with:

$a_{n}=\int_{0}^{1}\left(1-x+x^{2}-....+(-1)^{n-1}x^{n-1}\right)-\int_{0}^{1}\frac{1}{x+1}dx$
• Jul 5th 2008, 10:44 PM
CaptainBlack
Quote:

Originally Posted by galactus
Let $a_{n}=1-\frac{1}{2}+\frac{1}{3}-....+\frac{(-1)^{n-1}}{n}-ln(2)$

Prove $\sum_{n=1}^{\infty}a_{n}$ converges and find its sum.

This shouldn't be too bad. Just a neat series.

I started with:

$a_{n}=\int_{0}^{1}\left(1-x+x^{2}-....+(-1)^{n-1}x^{n-1}\right)-\int_{0}^{1}\frac{1}{x+1}dx$

1. The series converges by the Aternating series test.

2. Abels theorem tells us that:

If $\sum_0^{\infty} a_n$ converges, then:

$
\lim_{x \to 1^-}\sum_0^{\infty} a_n x^n = \sum_0^{\infty} a_n
$

Now:

$\ln(1+x)=\sum_1^{\infty} (-1)^{n-1}\frac{x^n}{n}$

for $x \in (-1,1]$

Hence with just a bit of manipulation we conclude that:

$\left[\sum_1^{\infty}\frac{(-1)^{n-1}}{n}\right]-\ln(2)=0$

RonL
• Jul 5th 2008, 11:22 PM
flyingsquirrel
Hello
Quote:

Originally Posted by CaptainBlack
1. The series converges by the Aternating series test.

2. Abels theorem tells us that:

If $\sum_0^{\infty} a_n$ converges, then:

$
\lim_{x \to 1^-}\sum_0^{\infty} a_n x^n = \sum_0^{\infty} a_n
$

Now:

$\ln(1+x)=\sum_1^{\infty} (-1)^{n-1}\frac{x^n}{n}$

for $x \in (-1,1]$

Hence with just a bit of manipulation we conclude that:

$\left[\sum_1^{\infty}\frac{(-1)^{n-1}}{n}\right]-\ln(2)=0$

Isn't the series that Galactus has to study $\sum_{n=1}^{\infty} \left[-\ln2+\sum_{k=1}^n\frac{(-1)^{k-1}}{k} \right]$ ?
• Jul 6th 2008, 06:04 AM
galactus
Well done to all who attempted.

Here is what I done.

Starting where I left off:

$a_{n}=\int_{0}^{1}\left(1-x+x^{2}-x^{3}+.....+(-1)^{n-1}x^{n-1}\right)dx-\int_{0}^{1}\frac{1}{1+x}dx$

$=\int_{0}^{1}\left(\frac{1+(-1)^{n-1}x^{n}}{1+x}\right)dx-\int_{0}^{1}\frac{1}{1+x}dx$

$=\int_{0}^{1}\frac{(-1)^{n-1}x^{n}}{1+x}dx$

Now, for any integer $N\geq 1$, we get:

$\sum_{n=1}^{\infty}a_{n}=\sum_{n=1}^{N}\int_{0}^{1 }\frac{(-1)^{n-1}x^{n}}{1+x}dx$

$=\int_{0}^{1}\frac{1}{1+x}\sum_{n=1}^{N}(-1)^{n-1}x^{n}dx$

$=\int_{0}^{1}\frac{x+(-1)^{N+1}x^{N+1}}{(1+x)^{2}}dx$

$=\int_{0}^{1}\frac{x}{(1+x)^{2}}dx+(-1)^{N+1}\int_{0}^{1}\frac{x^{N+1}}{(1+x)^{2}}dx$

Therefore,

$\left|\sum_{n=1}^{N}a_{n}-\int_{0}^{1}\frac{x}{(x+1)^{2}}dx\right|$

$=\int_{0}^{1}\frac{x^{N+1}}{(1+x)^{2}}dx\leq \int_{0}^{1}x^{N+1}dx=\frac{1}{N+2}$

Now, if we let $N\to {\infty}$ we can see that

$\sum_{n=1}^{\infty}a_{n}$ converges, and it has the sum

$\int_{0}^{1}\frac{x}{(1+x)^{2}}dx=\int_{0}^{1}\lef t(\frac{1}{1+x}-\frac{1}{(1+x)^{2}}\right)dx$

$=\boxed{ln(2)-\frac{1}{2}}$

Which, it would appear, is pretty much the same idea that Paul had.
• Jul 6th 2008, 06:58 AM
ThePerfectHacker
See this.