20. Convergence of Positive Series

e.1. The Ratio Test

The next test mimics the behavior of the geometric series.

The Ratio Test
Suppose ∑n=no∞an\displaystyle \sum_{n=n_o}^\infty a_n is a positive series and the limit of the ratio of successive terms is ρ=lim⁡n→∞an+1an\displaystyle \rho=\lim_{n\to\infty}\dfrac{a_{n+1}}{a_n}.

  1. If ρ<1\rho<1, then ∑n=no∞an\displaystyle \sum_{n=n_o}^\infty a_n is convergent.
  2. If ρ>1\rho>1, then ∑n=no∞an\displaystyle \sum_{n=n_o}^\infty a_n is divergent.

  ⟸\Longleftarrow Read this! It's easy.
  ⟸\Longleftarrow The proof requires the precise definition of the limit of a sequence.

If ρ=1\rho=1 the Ratio Test FAILS and says nothing about ∑n=no∞an\displaystyle \sum_{n=n_o}^\infty a_n.

Justification

For large nn, an+1an\dfrac{a_{n+1}}{a_n} is approximately ρ=lim⁡n→∞an+1an\displaystyle \rho=\lim_{n\to\infty}\dfrac{a_{n+1}}{a_n}. So pick some large NN. Then for n≥Nn \ge N, we have an≈aNρn−Na_n\approx a_N \rho^{n-N}. So we can approximate the tail of the series by ∑n=N∞an≈∑n=N∞aNρn−N \sum_{n=N}^\infty a_n \approx\sum_{n=N}^\infty a_N \rho^{n-N} This is a geometric series which converges to ∑n=N∞aNρn−N=aN1−ρ \sum_{n=N}^\infty a_N \rho^{n-N} =\dfrac{a_N}{1-\rho} provided ∣ρ∣<1|\rho| \lt 1 and diverges if ∣ρ∣≥1|\rho| \ge 1. So it is not surprizing that the original series also convegres or diverges when ρ\rho is less than or greater than 11 with the exception that the borderline case of ρ=1\rho=1 is indeterminate instead of divergent.

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Notice that the conclusions of the Ratio Test are the same as for a geometric series ∑n=no∞aρn\displaystyle \sum_{n=n_o}^\infty a\rho^n whose ratio is ρ\rho, except that the borderline case of ρ=1\rho=1 is indeterminate instead of divergent.

Determine if ∑n=0∞2nn!\displaystyle \sum_{n=0}^\infty \dfrac{2^n}{n!} is convergent or divergent.

We have an=2nn!a_n=\dfrac{2^n}{n!} and an+1=2n+1(n+1)!a_{n+1}=\dfrac{2^{n+1}}{(n+1)!}. So: (Notice how we divide by ana_n by multiplying by its reciprocal.) ρ=lim⁡n→∞an+1an=lim⁡n→∞2n+1(n+1)!⋅n!2n=lim⁡n→∞2n+1=0\begin{aligned} \rho&=\lim_{n\to\infty}\dfrac{a_{n+1}}{a_n} =\lim_{n\to\infty}\dfrac{2^{n+1}}{(n+1)!}\cdot\dfrac{n!}{2^n} \\ &=\lim_{n\to\infty}\dfrac{2}{n+1}=0 \end{aligned} Since ρ=0<1\rho=0 \lt 1, the series ∑n=0∞2nn!\displaystyle \sum_{n=0}^\infty \dfrac{2^n}{n!} converges.

n!(n+1)!=n⋯1(n+1)n⋯1=1n+1\dfrac{n!}{(n+1)!}=\dfrac{n\cdots1}{(n+1)n\cdots1}=\dfrac{1}{n+1}

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Determine if ∑n=0∞32n+12n+1\displaystyle \sum_{n=0}^\infty \dfrac{3^{2n+1}}{2n+1} is convergent or divergent.

We have an=32n+12n+1a_n=\dfrac{3^{2n+1}}{2n+1} and an+1=32(n+1)+12(n+1)+1=32n+32n+3a_{n+1}=\dfrac{3^{2(n+1)+1}}{2(n+1)+1}=\dfrac{3^{2n+3}}{2n+3}. So ρ=lim⁡n→∞an+1an=lim⁡n→∞32n+32n+3⋅2n+132n+1=lim⁡n→∞32(2n+1)2n+3=9\begin{aligned} \rho&=\lim_{n\to\infty}\dfrac{a_{n+1}}{a_n} =\lim_{n\to\infty}\dfrac{3^{2n+3}}{2n+3}\cdot\dfrac{2n+1}{3^{2n+1}} \\ &=\lim_{n\to\infty}\dfrac{3^2(2n+1)}{2n+3}=9 \end{aligned} Since ρ=9>1\rho=9>1, the series ∑n=0∞32n+12n+1\displaystyle \sum_{n=0}^\infty \dfrac{3^{2n+1}}{2n+1} diverges.

Determine if ∑n=0∞n2n2+1\displaystyle \sum_{n=0}^\infty \dfrac{n}{2n^2+1} is convergent or divergent.

We have an=n2n2+1a_n=\dfrac{n}{2n^2+1} and an+1=n+12(n+1)2+1a_{n+1}=\dfrac{n+1}{2(n+1)^2+1}. So ρ=lim⁡n→∞an+1an=lim⁡n→∞n+12(n+1)2+1⋅2n2+1n=lim⁡n→∞2n2+12(n+1)2+1⋅lim⁡n→∞n+1n=1⋅1=1\begin{aligned} \rho&=\lim_{n\to\infty}\dfrac{a_{n+1}}{a_n} =\lim_{n\to\infty}\dfrac{n+1}{2(n+1)^2+1}\cdot\dfrac{2n^2+1}{n} \\ &=\lim_{n\to\infty}\dfrac{2n^2+1}{2(n+1)^2+1}\cdot \lim_{n\to\infty}\dfrac{n+1}{n}=1\cdot1=1 \end{aligned} Since ρ=1\rho=1, the Ratio Test fails and we need to use some other test to determine the convergence.

One way is to use the Integral Test. The function f(n)=n2n2+1f(n)=\dfrac{n}{2n^2+1} is continuous, positive and decreasing and ∫0∞n2n2+1dn=14ln⁡(2n2+1)∣0∞=∞ \int_0^\infty \dfrac{n}{2n^2+1}\,dn =\left. \dfrac{1}{4}\ln(2n^2+1)\right|_0^\infty=\infty So ∑n=0∞n2n2+1\displaystyle \sum_{n=0}^\infty \dfrac{n}{2n^2+1} is divergent.

Another way is to use the Limit Comparison Test. Since an=n2n2+1≈12na_n=\dfrac{n}{2n^2+1}\approx\dfrac{1}{2n}, we compare to ∑n=0∞bn=∑n=0∞1n\displaystyle \sum_{n=0}^\infty b_n=\sum_{n=0}^\infty \dfrac{1}{n} which is the harmonic series which is divergent. So we compute the limit: lim⁡n→∞anbn=lim⁡n→∞n2n2+1n1=12 \lim_{n\to\infty}\dfrac{a_n}{b_n} =\lim_{n\to\infty}\dfrac{n}{2n^2+1}\dfrac{n}{1}=\dfrac{1}{2} Since 0<12<∞0 \lt \dfrac{1}{2} \lt \infty we know ∑n=0∞n2n2+1\displaystyle \sum_{n=0}^\infty \dfrac{n}{2n^2+1} is also divergent.

Determine if ∑n=1∞n3n\displaystyle \sum_{n=1}^\infty \dfrac{n}{3^n} is convergent or divergent.

Answer

∑n=1∞n3n\displaystyle \sum_{n=1}^\infty \dfrac{n}{3^n} is convergent.

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Solution

We have an=n3na_n=\dfrac{n}{3^n} and an+1=n+13n+1a_{n+1}=\dfrac{n+1}{3^{n+1}}. So ρ=lim⁡n→∞an+1an=lim⁡n→∞n+13n+1⋅3nn=lim⁡n→∞n+13n=13\begin{aligned} \rho=\lim_{n\to\infty}\dfrac{a_{n+1}}{a_n} =\lim_{n\to\infty}\dfrac{n+1}{3^{n+1}}\cdot\dfrac{3^n}{n} \\ =\lim_{n\to\infty}\dfrac{n+1}{3n}=\dfrac{1}{3} \end{aligned} Since ρ=13<1\rho=\dfrac{1}{3}< 1, the series ∑n=0∞n3n\displaystyle \sum_{n=0}^\infty \dfrac{n}{3^n} converges.

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Determine if ∑n=1∞2nn!(n+1)!\displaystyle \sum_{n=1}^\infty \dfrac{2^nn!}{(n+1)!} is convergent or divergent.

Answer

∑n=1∞2nn!(n+1)!\displaystyle \sum_{n=1}^\infty \dfrac{2^nn!}{(n+1)!} is divergent.

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Solution

We have an=2nn!(n+1)!a_n=\dfrac{2^nn!}{(n+1)!} and an+1=2n+1(n+1)!(n+2)!a_{n+1}=\dfrac{2^{n+1}(n+1)!}{(n+2)!}. So ρ=lim⁡n→∞an+1an=lim⁡n→∞2n+1(n+1)!(n+2)!⋅(n+1)!2nn!=lim⁡n→∞2(n+1)n+2=2\begin{aligned} \rho&=\lim_{n\to\infty}\dfrac{a_{n+1}}{a_n} =\lim_{n\to\infty}\dfrac{2^{n+1}(n+1)!}{(n+2)!}\cdot\dfrac{(n+1)!}{2^nn!} \\ &=\lim_{n\to\infty}\dfrac{2(n+1)}{n+2}=2 \end{aligned} Since ρ=2>1\rho=2 \gt 1, the series ∑n=1∞n!2n(n+1)!\displaystyle \sum_{n=1}^\infty \dfrac{n!2^n}{(n+1)!} diverges by the ratio test.

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Remark

Notice that the Divergence Test will also work, since lim⁡n→∞2nn!(n+1)!=lim⁡n→∞2nn+1=∞≠0 \lim_{n\to\infty} \dfrac{2^nn!}{(n+1)!} =\lim_{n\to\infty} \dfrac{2^n}{n+1}=\infty \ne 0

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