# Strategies to Test an Infinite Series for Convergence

## Key Questions

• 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.

• There is no general method of determining the test you should use to check the convergence of a series.

• For series where the general term has exponents of $n$, it's useful to use the root test (also known as Cauchy's test).
Example 1: Power Series
The definition of the convergence radius of the of a power series comes from the Cauchy test (however, the actual computation is usually done with the following test).

• Generally, the computation of the ratio test (also known as d'Alebert's test) is easier than the computation of the root test.
Example 2: Inverse Factorial
For the series sum_(n=1)^(oo) 1/(n!) the d'Alembert's test gives us:
lim_(n to oo) |1/((n+1)!)|/|1/(n!)| = lim_(n to oo) |n!|/(|(n+1)!|) = lim_(n to oo) |(n!)/((n+1)n!)| = lim_(n to oo) |1/(n+1)| = 0
So the series is convergent.

• If you know the result of the improper integral of the function $f \left(x\right)$ such that $f \left(n\right) = {a}_{n}$, where ${a}_{n}$ is the general term of the series being analyzed, then it might be a good idea to use the integral test.
Example 3: A proof for the Harmonic Series.
Knowing that the improper integral ${\int}_{1}^{\infty} \frac{1}{x} \mathrm{dx}$ is divergent (it's easy to check) implies that the harmonic series ${\sum}_{n = 1}^{\infty} \frac{1}{n}$ diverges.

• Comparision tests are only useful if you know an appropriate series to compare the one you're analyzing to. However, they can be very powerful.
Example 4: Hyperharmonic Series
The series of the form ${\sum}_{n = 1}^{\infty} \frac{1}{{n}^{p}}$ are called hyperharmonic series or $p$-series. If you can show that the series ${\sum}_{n = 1}^{\infty} \frac{1}{{n}^{1 + \epsilon}}$ converges, for some small, positive value of $\epsilon$, than any $p$-series such that $p > 1 + \epsilon$ converges.