# Thread: another series...not fibonacci

1. ## another series...not fibonacci

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

Find the sum of the infinite series:

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

2. 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:

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

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

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

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

Thus, we have:

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

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

Thus we can see that:

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

Was that the way you did it galactus?

3. 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.

4. 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

5. Here's something you may find interesting then.

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

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

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

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

and on and on.

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

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

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

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

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

$\displaystyle 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.

6. Originally Posted by galactus
Here's something you may find interesting then.

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

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

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

7. 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?.

8. 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 $\displaystyle \sin ( \pi x )$

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

Let me just post a real cool infinite summation:
$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2+a^2} = \frac{1}{2a}\mbox{coth} \pi a - \frac{1}{2a^2} \mbox{ for }0<a<1$.
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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.