Strategies to Test an Infinite Series for Convergence
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Key Questions

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(x)# such that#f(n)=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^(oo) 1/x dx# is divergent (it's easy to check) implies that the harmonic series#sum_(n=1)^(oo) 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)^(oo) 1/(n^p)# are called hyperharmonic series or#p# series. If you can show that the series#sum_(n=1)^(oo) 1/(n^(1+epsilon))# converges, for some small, positive value of#epsilon# , than any#p# series such that#p>1 + epsilon# converges.


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 ballpark 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}/{11/e^{n+1}}={0+1}/{10}=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}/{1e^{n+1}}# also converges since negation does not affect the convergence of the series.
I hope that this was helpful.
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Tests of Convergence / Divergence

1Geometric Series

2Nth Term Test for Divergence of an Infinite Series

3Direct Comparison Test for Convergence of an Infinite Series

4Ratio Test for Convergence of an Infinite Series

5Integral Test for Convergence of an Infinite Series

6Limit Comparison Test for Convergence of an Infinite Series

7Alternating Series Test (Leibniz's Theorem) for Convergence of an Infinite Series

8Infinite Sequences

9Root Test for for Convergence of an Infinite Series

10Infinite Series

11Strategies to Test an Infinite Series for Convergence

12Harmonic Series

13Indeterminate Forms and de L'hospital's Rule

14Partial Sums of Infinite Series