1. ## Random Series Question

This is just a random series question I was curious about. I know that if $\sum {a_n}$ and $\sum {b_n}$ are both convergent then,

$\sum {a_n} \pm \sum {b_n}=\sum ({a_n}\pm {b_n})$

But, I was curious, can this be done if either one of the two series is convergent and the other divergent or if both are divergent. To give a specific example, I was looking at this series:

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

One can obviously prove it's divergence by the integral test, but I was wondering if it would also be alright to say something like this:

$\sum_{n=2}^{\infty }\frac{n^2-1}{2n^3}}=\sum_{n=2}^{\infty }(\frac{n^2}{2n^3}-\frac{1}{2n^3})=\sum_{n=2}^{\infty }\frac{n^2}{2n^3}-\sum_{n=2}^{\infty }\frac{1}{2n^3}=\frac{1}{2}\sum_{n=2}^{\infty }\frac{1}{n}-\frac{1}{2}\sum_{n=2}^{\infty }\frac{1}{n^3}$

Thus, the series diverges since the first sum is divergent and the second sum is convergent (both by the p-test).

So, I'm basically asking two questions.

Does this operation described above work generally when you are dealing with divergent series or cases where one series diverges and one converges?

And if it is not generally true, does the manipulation described above work in the specific example I gave or was it just a nice coincidence that it worked on this specific example. Also, if it does work sometimes and not other times, why would this be so?

Thanks.

2. ## Re: Random Series Question

If you can use algebra to convert one series into the sum of 2 series where one diverges and one converges then you can always conclude divergence as div+con = div, div - con = div, con - div = div

Just be sure the algebra in converting the terms is valid