Strategies to Test an Infinite Series for Convergence
Key Questions
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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.
I hope that this was helpful.
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There is no general method of determining the test you should use to check the convergence of a series.
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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 seriessum_(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 thatf(n)=a_n , wherea_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 integralint_1^(oo) 1/x dx is divergent (it's easy to check) implies that the harmonic seriessum_(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 formsum_(n=1)^(oo) 1/(n^p) are called hyperharmonic series orp -series. If you can show that the seriessum_(n=1)^(oo) 1/(n^(1+epsilon)) converges, for some small, positive value ofepsilon , than anyp -series such thatp>1 + epsilon converges.
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Questions
Tests of Convergence / Divergence
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Geometric Series
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Nth Term Test for Divergence of an Infinite Series
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Direct Comparison Test for Convergence of an Infinite Series
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Ratio Test for Convergence of an Infinite Series
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Integral Test for Convergence of an Infinite Series
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Limit Comparison Test for Convergence of an Infinite Series
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Alternating Series Test (Leibniz's Theorem) for Convergence of an Infinite Series
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Infinite Sequences
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Root Test for for Convergence of an Infinite Series
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Infinite Series
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Strategies to Test an Infinite Series for Convergence
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Harmonic Series
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Indeterminate Forms and de L'hospital's Rule
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Partial Sums of Infinite Series