# sum of alternating series

• December 30th 2013, 11:29 AM
hedi
sum of alternating series
Hi,
I am trying to prone that the sum of the following series is log3.

1+1/2-2/3+1/4+1/5-2/6+1/7+1/8-2/9+1/10+1/11-2/12...

• December 30th 2013, 02:59 PM
chiro
Re: sum of alternating series
Hey hedi.

Hint: Look at the taylor series expanded around a = 1 (i.e. taylor series for log(1+x)).
• December 30th 2013, 03:06 PM
hedi
Re: sum of alternating series
This gives the harmonic alternating series,it is not so helpfull here.
• December 30th 2013, 03:21 PM
chiro
Re: sum of alternating series
Can you collect terms together to show that both are the same series?
• December 30th 2013, 03:25 PM
romsek
Re: sum of alternating series
Quote:

Originally Posted by hedi
Hi,
I am trying to prone that the sum of the following series is log3.

1+1/2-2/3+1/4+1/5-2/6+1/7+1/8-2/9+1/10+1/11-2/12...

I think this is the continued fraction form of the natural log. Take a look at the wiki page on continued fractions here or here
• December 31st 2013, 05:43 AM
SlipEternal
Re: sum of alternating series
Quote:

Originally Posted by hedi
Hi,
I am trying to prone that the sum of the following series is log3.

1+1/2-2/3+1/4+1/5-2/6+1/7+1/8-2/9+1/10+1/11-2/12...

Let $H_n$ be the $n$-th Harmonic number (the sum of the first $n$ terms of the Harmonic series). Consider partial sums of three terms of your series at a time. Show that $S_n = H_{3n}-H_n$ gives the correct partial sums for your series. So, your series is given by $\lim_{n \to \infty}S_n$.

Next, use common inequalities for the Harmonic numbers. For example, Young (1991) showed that $\dfrac{1}{2n+2} < H_n - \ln n - \gamma < \dfrac{1}{2n}$. So,

$\dfrac{1}{6n+2} < H_{3n} - \ln (3n) - \gamma < \dfrac{1}{6n}$ and $-\dfrac{1}{2n} < \gamma + \ln n - H_n < -\dfrac{1}{2n+2}$

$\dfrac{1}{6n+2} - \dfrac{1}{2n} < H_{3n} + \gamma + \ln n - \ln (3n) - \gamma - H_n < \dfrac{1}{6n} - \dfrac{1}{2n+2}$

Simplifying gives:

$-\dfrac{2n+1}{2n(3n+1)} < S_n - \ln 3 < \dfrac{1-2n}{6n(n+1)}$

Finally, use the Squeeze Theorem to show that $\lim_{n \to \infty} (S_n - \ln 3) = 0$, so $\lim_{n \to \infty} S_n = \ln 3$
• December 31st 2013, 07:31 AM
SlipEternal
Re: sum of alternating series
Btw, to show that $S_n = H_{3n} - H_n$, you can write your series as follows:

\begin{align*}1+\frac{1}{2} -\frac{2}{3} +\cdots + \frac{1}{3n-2} + \frac{1}{3n-1} - \frac{2}{3n} & = \sum_{k= 1}^n \left(\dfrac{1}{3k-2} + \dfrac{1}{3k-1} -\dfrac{2}{3k}\right) \\ & = \sum_{k= 1}^n \left(\dfrac{1}{3k-2} + \dfrac{1}{3k-1} + \dfrac{1}{3k} -\dfrac{3}{3k}\right) \\ & = \sum_{k= 1}^n \left(\dfrac{1}{3k-2} + \dfrac{1}{3k-1} + \dfrac{1}{3k}\right) -\sum_{k=1}^n\dfrac{1}{k} \\ & = \sum_{k=1}^{3n}\dfrac{1}{k} - \sum_{k=1}^n \dfrac{1}{k} \\ & = H_{3n} - H_n\end{align*}
• December 31st 2013, 08:03 AM
hedi
Re: sum of alternating series
Thanks to both of you so much.
• December 31st 2013, 11:50 AM
romsek
Re: sum of alternating series
I'm curious what course (if any) this would be learned in. I've had a fair bit of math and never ran into a detailed analysis of the Harmonic series like this.

Is this something you just naturally come across post-grad in the course of solving other problems?
• December 31st 2013, 01:06 PM
SlipEternal
Re: sum of alternating series
Quote:

Originally Posted by romsek
I'm curious what course (if any) this would be learned in. I've had a fair bit of math and never ran into a detailed analysis of the Harmonic series like this.

Is this something you just naturally come across post-grad in the course of solving other problems?

Are you asking hedi or me? If you are asking me, then this comes about from examining Euler's constant, which is $\gamma = \lim_{n \to \infty} H_n - \ln n$. Young was showing how quickly the sequence $H_n - \ln n$ converges to $\gamma$.
• December 31st 2013, 01:13 PM
hedi
Re: sum of alternating series
This question comes from a course in problem solving.
• December 31st 2013, 01:46 PM
SlipEternal
Re: sum of alternating series
Actually, I just thought of a more direct way to show that the series converges to $\ln 3$. By definition, $\gamma := \lim_{n \to \infty} (H_n - \ln n)$, so

\begin{align*}0 & = \gamma - \gamma \\ & = \lim_{n \to \infty} \left(H_{3n} - \ln (3n) \right) - \lim_{n \to \infty} \left( H_n - \ln n \right) \\ & = \lim_{n \to \infty} \left( H_{3n} - H_n - \ln 3 \right)\end{align*}
• January 1st 2014, 04:14 PM
Prove It
Re: sum of alternating series
This is not an alternating series...