# Verifying if series is convergent

1.Is this series convergent? If so, how do we know and what is the exact sum? $\sum_{n=2}^{\infty} \frac{6}{n^2 + n - 2}$

So, can I use the comparison test? Would it be right to say that the above series is less than $\frac{6}{n^2}$, which converges by the rules of the "P-series" (where p > 1), and so it must, therefore, also converge? As for finding the exact sum, would I write out a sequence of partial sums, and see what it appears to converge to?

1. Is this series convergent? If so, how do we know and what is the exact sum? $\sum_{n=1}^{\infty} (3^{-2n})\frac{4^n}{2}$

For this one, I am at a loss on how to begin ... Do I need to use the alternating series test? If so, what would that look like? Also, not sure on how to find the exact sum here ...

Thank you in advance!

As for the first, your argument is correct, and it follows that it is convergent. As a hint for finding the sum, use partial fractions to get $$\frac{6}{n^2+n-2} = \frac{2}{n-1} - \frac{2}{n+2}$$ and note that a type of telescoping happens.

For the second, you have that $$(3^{-2n})\frac{4^n}{2} = (3^{-2})^n \frac{4^n}{2} = \frac{1}{9^n}\frac{4^n}{2} = \frac{1}{2}\left(\frac{4}{9}\right)^n,$$ and use geometric series.

• So, when I expanded the terms out to try to get the closed form, it looked like everything but $\frac{1}{3}$ cancels out. But, that doesn't seem right ...2[$(\frac{-1}{4} + \frac{1}{3}) + (\frac{-1}{5} + \frac{1}{4}) + (\frac{-1}{6}+\frac{1}{5}) +...$] – LaSpana101 Nov 14 '16 at 5:15
• The difference between $n+2$ and $n-1$ is $3$. Your terms should be $$\left(\frac{2}{1} - \frac{2}{4}\right) + \left(\frac{2}{2} - \frac{2}{5}\right)+\left(\frac{2}{3} - \frac{2}{6}\right)+\left(\frac{2}{4} - \frac{2}{7}\right)...$$ so everything cancels except $$\frac{2}{1} + \frac{2}{2} + \frac{2}{3} = \frac{11}{3}.$$ – Eff Nov 14 '16 at 6:27

By comparison test, since $n^2 + n -2 \geq n^2$ whenever $n\geq 2$. So the general term is greater than $0$ but smaller than $\frac 6{n^2}$, so it converges.

To find its limit, first by partial fraction decomposition:

$$\frac 6{n^2 +n-2} = \frac A{n+2} +\frac B{n-1}$$

solving gives: $\frac {-2}{n+2} +\frac 2{n-1}$

It is telescoping, try to find a closed form for the partial sum. Then take limit.

For the second series, try to use ratio test.