another series...not fibonacci

**galactus** · July 14th 2007, 10:18 AM

Hey Soroban, here's one you may like. Or anyone else, by all means.

Find the sum of the infinite series:

$\sum_{n=2}^{\infty}\frac{1}{n^{2}+3n-4}$

**Jhevon** · July 14th 2007, 02:50 PM

Originally Posted by galactus

Hey Soroban, here's one you may like. Or anyone else, by all means.

Find the sum of the infinite series:

$\sum_{n=2}^{\infty}\frac{1}{n^{2}+3n-4}$

$\sum_{n = 2}^{ \infty} \frac {1}{(n + 4)(n - 1)}$

splitting the last fraction by partial fractions, we find that the sum is actually:

$\sum_{n = 2}^{ \infty} \left( \frac { \frac {1}{5}}{n - 1} - \frac { \frac {1}{5}}{n + 4} \right)$

Thus, we have:

$\frac {1}{5} \sum_{n = 2}^{ \infty} \left( \frac {1}{n - 1} - \frac {1}{n + 4} \right) = \frac {1}{5} \left( 1 - \frac {1}{6} \right.$ $\left. + \frac {1}{2} - \frac {1}{7} + \frac {1}{3} - \frac {1}{8} + \frac {1}{4} - \frac {1}{9} + \frac {1}{5} - \frac {1}{10} + \frac {1}{6} - \frac {1}{11} ... \right)$

After writing out a few terms, it does not take long to realize that the $- \frac {1}{n + 4}$ will cancel all the terms from $\frac {1}{n - 1}$ that are $\geq \frac {1}{6}$

Thus we can see that:

$\frac {1}{5} \sum_{n = 2}^{ \infty} \left( \frac {1}{n - 1} - \frac {1}{n + 4} \right) = \frac {1}{5} \left( 1 + \frac {1}{2} + \frac {1}{3} + \frac {1}{4} + \frac {1}{5} \right) = \frac {137}{300}$

Was that the way you did it galactus?

**galactus** · July 14th 2007, 03:49 PM

Yes, that's exactly how I approached it. It wasn't too bad. Just fun.

I like these converging infinite series. Sometimes they're challenging and result in a cool sum.

**Jhevon** · July 14th 2007, 03:51 PM

Originally Posted by galactus

I like these converging infinite series. Sometimes they're challenging and result in a cool sum.

yeah, i like the ones that end up having pi in the answer

in general i'm not good with series though

**galactus** · July 14th 2007, 04:22 PM

Here's something you may find interesting then.

You are probably familiar with Euler's Basel problem. You know

$\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{{\pi}^{2} }{6}$

$\sum_{n=1}^{\infty}\frac{1}{n^{4}}=\frac{{\pi}^{4} }{90}$

$\sum_{n=1}^{\infty}\frac{1}{n^{6}}=\frac{{\pi}^{6} }{945}$

and on and on.

Well, Euler found a general formula for the sum of the reciprocals of the EVEN powers.

$\sum_{n=1}^{\infty}\frac{1}{n^{2k}}=\frac{(-1)^{k-1}B_{2k}(2\pi)^{2k}}{2(2k)!}$

So if you wanted $\sum_{n=1}^{\infty}\frac{1}{n^{20}}$

You could use the formula and get $\frac{174611{\pi}^{20}}{1531329465290625}$

The $B_{2k}$ is the even Bernoulli numbers. They are tied in here also. Isn't it cool how this stuff ties together.

$B_{2}=\frac{1}{6}, \;\ B_{4}=\frac{-1}{30}, \;\ B_{6}=\frac{1}{42}, etc, \;\$

The Bernoulli numbers are an interesting thing to research if you're unfamiliar with them. Just a thought.

**Jhevon** · July 14th 2007, 04:40 PM

Originally Posted by galactus

Here's something you may find interesting then.

You are probably familiar with Euler's Basel problem. You know

$\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{{\pi}^{2} }{6}$

I took an advanced calculus class last semester and the professor actually derived this.

Well, Euler found a general formula for the sum of the reciprocals of the EVEN powers.

$\sum_{n=1}^{\infty}\frac{1}{n^{2k}}=\frac{(-1)^{k-1}B_{2k}(2\pi)^{2k}}{2(2k)!}$

i think he mentioned this formula as well

The Bernoulli numbers are an interesting thing to research if you're unfamiliar with them. Just a thought.

I plan on looking up Bernoulli numbers one day (i'm one of those procrastinators, so chances are i won't look them up today even if i said i would). I don't know anything about them other than the fact they are used in this formula.

**galactus** · July 14th 2007, 04:43 PM

There's a bunch of different methods of deriving the Basel problem. The most popular using the Taylor expansion for sine. Is that the one your professor used?.

**Jhevon** · July 14th 2007, 04:50 PM

Originally Posted by galactus

There's a bunch of different methods of deriving the Basel problem. The most popular using the Taylor expansion for sine. Is that the one your professor used?.

yes. i believe he used the Taylor series expansion for $\sin ( \pi x )$

**ThePerfectHacker** · July 14th 2007, 05:48 PM

I can derive this with Complex Analysis if anyone is interested.

Let me just post a real cool infinite summation:
$\sum_{n=1}^{\infty} \frac{1}{n^2+a^2} = \frac{1}{2a}\mbox{coth} \pi a - \frac{1}{2a^2} \mbox{ for }0<a<1$ .
-----------------------------
Leonard Euler's immortal approach was not formal.

His idea was that a function can be factored by its zero's like a polynomial. Though his idea was informal it attracted attention. The general theorem is known as the Weierstrauss Factorization Theorem.

Off Topic: How is "Weierstrauss" pronounced properly, without an Americanized accent. Because I hated how my professor pronounced his name, it is so....so Americanized, if you know what I mean.

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