Using Integral methods to find a summation of series (infinite)

[Amad27]

Hi, let's take the sum:

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{9n^2 + 3n - 2}$

$\implies 9n^2 + 3n - 2 = 9n^2 + 6n - 3n - 2 = 3n(3n + 2) - (3n + 2) = (3n - 1)(3n - 2)$

The simplest way would be to use partial fractions, and then convert this into a telescoping series. Which makes the sum extremely simple, but I am looking for a way in, which I could use integrals, perhaps a definite integral in, which I could derive this sum. Any ideas?

The $f(n)$ is $f(n) = \frac{1}{(3n-1)(3n-2)}$

$f'(n) = \frac{-9(2n-1)}{(9n^2 - 9n + 2)^2}$

$\displaystyle \int_{n}^{n+1} \frac{-9(2x-1)}{(9x^2 - 9x + 2)^2} \,dx$ does not work.

I was thinking of making a infinite sum partition of an improper integral part of, which the result would the required sum: $\displaystyle \sum_{n=1}^{\infty}\frac{1}{9n^2 + 3n - 2}$

 

[alyafey22]

Hi, let's take the sum:

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{9n^2 + 3n - 2}$

$\implies 9n^2 + 3n - 2 = 9n^2 + 6n - 3n - 2 = 3n(3n + 2) - (3n + 2) = (3n - 1)(3n - 2)$

The simplest way would be to use partial fractions, and then convert this into a telescoping series. Which makes the sum extremely simple,

You have made a mistake it should be $(3n+2)(3n-1)$.

 

[Amad27]

You have made a mistake it should be $(3n+2)(3n-1)$.

that is indeed what I meant, you can see that I meant that by seeing the factors, it was a typo. Thanks!

But so, is there a way for doing this using integration.

For some reason integration and its link to series interests me a lot, so that is why I am trying to find this out!

Thanks!

 

[alyafey22]

that is indeed what I meant, you can see that I meant that by seeing the factors, it was a typo. Thanks!

But so, is there a way for doing this using integration.

For some reason integration and its link to series interests me a lot, so that is why I am trying to find this out!

Thanks!

You already did it correctly in this http://mathhelpboards.com/calculus-10/interchanging-summation-integrals-13033.html#post61800.