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?
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.
I took an advanced calculus class last semester and the professor actually derived this.
i think he mentioned this formula as well
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 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.The Bernoulli numbers are an interesting thing to research if you're unfamiliar with them. Just a thought.
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.