# Thread: prove convergence and find sum

1. ## prove convergence and find sum

Let $\displaystyle a_{n}=1-\frac{1}{2}+\frac{1}{3}-....+\frac{(-1)^{n-1}}{n}-ln(2)$

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

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

I started with:

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

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

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

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

I started with:

$\displaystyle a_{n}=\int_{0}^{1}\left(1-x+x^{2}-....+(-1)^{n-1}x^{n-1}\right)-\int_{0}^{1}\frac{1}{x+1}dx$
1. The series converges by the Aternating series test.

2. Abels theorem tells us that:

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

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

Now:

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

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

Hence with just a bit of manipulation we conclude that:

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

RonL

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

2. Abels theorem tells us that:

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

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

Now:

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

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

Hence with just a bit of manipulation we conclude that:

$\displaystyle \left[\sum_1^{\infty}\frac{(-1)^{n-1}}{n}\right]-\ln(2)=0$
Isn't the series that Galactus has to study $\displaystyle \sum_{n=1}^{\infty} \left[-\ln2+\sum_{k=1}^n\frac{(-1)^{k-1}}{k} \right]$ ?

4. Well done to all who attempted.

Here is what I done.

Starting where I left off:

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

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

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

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

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

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

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

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

Therefore,

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

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

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

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

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

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

Which, it would appear, is pretty much the same idea that Paul had.

5. See this.