Question

If I want to test the series `sum_(n=1)^oo(n^2+2^n)/(1-e^(n+1))` for convergence, what would be the best test to use and why?

Answer 1

By pulling out the negative sign so that `a_n ge0`, let

`-sum_{n=1}^inftya_n=-sum_{n=1}^infty{n^2+2^n}/{e^{n+1}-1}`.

I would use Limit Comparison Test since we can make a ball-park estimate of the series by only looking at the dominant terms on the numerator and the denominator. This series can be compared to

`sum_{n=1}^inftyb_n=sum_{n=1}^infty{2^n}/{e^{n+1}}=sum_{n=1}^infty1/e(2/e)^n`,

which is a convergent geometric series with `|r|=|2/e|<1`.

Let us make sure that they are comparable.

`lim_{n to infty}{a_n}/{b_n}=lim_{n to infty}{{n^2+2^n}/{e^{n+1}-1}}/{{2^n}/{e^{n+1}}}`

`=lim_{n to infty}{{n^2}/{2^n}+1}/{1-1/e^{n+1}}={0+1}/{1-0}=1 < infty`

(Note: `lim_{n to infty}{n^2}/2^n=0` by applying l'Hopital's Rule twice.)

So, `sum_{n=1}^infty a_n` converges by Limit Comparison Test.

Hence, `sum_{n=1}^infty{n^2+2^n}/{1-e^{n+1}}` also converges since negation does not affect the convergence of the series.

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I hope that this was helpful.