# Simple series

• Jan 16th 2011, 10:20 PM
chiph588@
Simple series
I'm not sure if this is worthy of the challenge section, but here goes.

Without using Taylor series/generating functions, show $\displaystyle \displaystyle \sum_{n=1}^\infty \frac{(-1)^n}n = -\log2$.
• Jan 16th 2011, 10:41 PM
Prove It
The MacLaurin Series for $\displaystyle \displaystyle \ln{(1+x)} = \sum_{n = 1}^{\infty}\frac{(-1)^{n+1}x^n}{n}$ for $\displaystyle \displaystyle -1 < x \leq 1$.

So $\displaystyle \displaystyle \ln{2} = \ln{(1 + 1)} = \sum_{n = 1}^{\infty}\frac{(-1)^{n+1}\cdot 1^n}{n} = \sum_{n = 1}^{\infty}\frac{(-1)^{n+1}}{n}$.

And finally $\displaystyle \displaystyle -\ln{2} = \frac{\ln{2}}{-1} = \frac{\sum_{n = 1}^{\infty}\frac{(-1)^{n+1}}{n}}{-1} = \sum_{n = 1}^{\infty}\frac{(-1)^n}{n}$.

Edit: Oops, didn't read the question fully stating not to use Taylor series hahaha. Ah well, it stays because it's beautiful ;)
• Jan 16th 2011, 10:59 PM
Prove It
Here's another...

$\displaystyle \displaystyle \ln{2} = \int_1^2{\frac{1}{x}\,dx}$.

Substitute $\displaystyle \displaystyle x = 1 + t$ so that $\displaystyle \displaystyle dx = dt$. When $\displaystyle \displaystyle x= 1, t = 0$ and when $\displaystyle \displaystyle x = 2, t = 1$ and the integral becomes

$\displaystyle \displaystyle \int_0^1{\frac{1}{1 + t}\,dt}$

$\displaystyle \displaystyle = \int_0^1{\frac{1}{1 - (-t)}\,dt}$

$\displaystyle \displaystyle = \int_0^1{\sum_{n = 0}^{\infty}(-t)^n\,dt}$ provided $\displaystyle \displaystyle |-t| < 1 \implies |t| < 1$

$\displaystyle \displaystyle = \sum_{n = 0}^{\infty}{\int_0^1{(-t)^n\,dt}}$

$\displaystyle \displaystyle = \sum_{n = 0}^{\infty}{\lim_{\epsilon \to 1}\int_0^{\epsilon}{(-t)^n\,dt}}$.

The rest should follow.
• Jan 16th 2011, 11:07 PM
TheCoffeeMachine
Quote:

$\displaystyle \displaystyle = \int_0^1{\frac{1}{1 - (-t)}\,dt}$

$\displaystyle \displaystyle = \int_0^1{\sum_{n = 0}^{\infty}(-t)^n\,dt}$ provided $\displaystyle \displaystyle |-t| < 1 \implies |t| < 1$
Isn't $\displaystyle \displaystyle \frac{1}{1+t} = \sum_{n\ge 0}(-1)^nt^n$ a Taylor series?
• Jan 16th 2011, 11:33 PM
FernandoRevilla
Hints

(i) $\displaystyle \displaystyle\sum_{k=1}^n{\dfrac{1}{k}}=\log n+C+o(1)\;\textrm{as}\;n\rightarrow{+\infty}$

(ii) The following sequence is convergent

$\displaystyle S_m=\displaystyle\sum_{i=1}^m{\dfrac{(-1)^k}{k}}$

(iii) Express

$\displaystyle S_{2n}=\ldots=\displaystyle\sum_{k=1}^n{\dfrac{1}{ k}}-\displaystyle\sum_{k=1}^{2n}{\dfrac{1}{k}}$

(iv) Apply (i) to prove that

$\displaystyle \displaystyle\lim_{n \to{+}\infty}{S_{2n}}=-\log 2$

which is exactly the sum of the series.

Fernando Revilla
• Jan 16th 2011, 11:52 PM
Prove It
Quote:

Originally Posted by TheCoffeeMachine
Isn't $\displaystyle \displaystyle \frac{1}{1+t} = \sum_{n\ge 0}(-1)^nt^n$ a Taylor series?

No, it's a Geometric Series :P
• Jan 17th 2011, 08:43 AM
chiph588@
Quote:

Originally Posted by FernandoRevilla

(ii) The following sequence is convergent

$\displaystyle S_m=\displaystyle\sum_{i=1}^m{\dfrac{(-1)^k}{k}}$

(iii) Express

$\displaystyle S_{2n}=\ldots=\displaystyle\sum_{k=1}^n{\dfrac{1}{ k}}-\displaystyle\sum_{k=1}^{2n}{\dfrac{1}{k}}$

You've got the right idea, but this step is not correct.

Quote:

Originally Posted by Prove It
No, it's a Geometric Series :P

It's also a Taylor series too!
• Jan 17th 2011, 09:28 AM
FernandoRevilla
Quote:

Originally Posted by chiph588@
You've got the right idea, but this step is not correct.

It is correct.

Hint : Express

$\displaystyle S_{2n}=\displaystyle\sum_{k=1}^n{\dfrac{1}{2k}}-\displaystyle\sum_{k=1}^n{\dfrac{1}{2k-1}}$

Fernando Revilla