# 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?

Oct 19, 2014

By pulling out the negative sign so that ${a}_{n} \ge 0$, let

$- {\sum}_{n = 1}^{\infty} {a}_{n} = - {\sum}_{n = 1}^{\infty} \frac{{n}^{2} + {2}^{n}}{{e}^{n + 1} - 1}$.

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

${\sum}_{n = 1}^{\infty} {b}_{n} = {\sum}_{n = 1}^{\infty} \frac{{2}^{n}}{{e}^{n + 1}} = {\sum}_{n = 1}^{\infty} \frac{1}{e} {\left(\frac{2}{e}\right)}^{n}$,

which is a convergent geometric series with $| r | = | \frac{2}{e} | < 1$.

Let us make sure that they are comparable.

${\lim}_{n \to \infty} \frac{{a}_{n}}{{b}_{n}} = {\lim}_{n \to \infty} \frac{\frac{{n}^{2} + {2}^{n}}{{e}^{n + 1} - 1}}{\frac{{2}^{n}}{{e}^{n + 1}}}$

$= {\lim}_{n \to \infty} \frac{\frac{{n}^{2}}{{2}^{n}} + 1}{1 - \frac{1}{e} ^ \left\{n + 1\right\}} = \frac{0 + 1}{1 - 0} = 1 < \infty$

(Note: ${\lim}_{n \to \infty} \frac{{n}^{2}}{2} ^ n = 0$ by applying l'Hopital's Rule twice.)

So, ${\sum}_{n = 1}^{\infty} {a}_{n}$ converges by Limit Comparison Test.

Hence, ${\sum}_{n = 1}^{\infty} \frac{{n}^{2} + {2}^{n}}{1 - {e}^{n + 1}}$ also converges since negation does not affect the convergence of the series.

I hope that this was helpful.