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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!

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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.

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  • $\begingroup$ 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[$(\frac{-1}{4} + \frac{1}{3}) + (\frac{-1}{5} + \frac{1}{4}) + (\frac{-1}{6}+\frac{1}{5}) +...$] $\endgroup$ – LaSpana101 Nov 14 '16 at 5:15
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    $\begingroup$ 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 $$\frac{2}{1} + \frac{2}{2} + \frac{2}{3} = \frac{11}{3}.$$ $\endgroup$ – Eff Nov 14 '16 at 6:27
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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.

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