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Exercises: Series Convergence

Theory

Determine whether each series converges or diverges.
If it converges, find its value.

Exercises

Exercise 1
∑n=1∞7(−16)n−1\sum_{n=1}^{\infty} 7\left(-\frac16\right)^{n-1}
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Solution

Geometric series with a=7a=7, r=−16r=-\frac16, |r|<1.

S=a1−r=71+16=6.S=\frac{a}{1-r}=\frac{7}{1+\frac16}=6.

Conclusion: Convergent to 66.

Exercise 2
∑n=1∞2(4n−3)(4n+1)\sum_{n=1}^{\infty} \frac{2}{(4n-3)(4n+1)}
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Solution

Partial fractions:

2(4n−3)(4n+1)=12 ⁣(14n−3−14n+1).\frac{2}{(4n-3)(4n+1)}=\frac12\!\left(\frac{1}{4n-3}-\frac{1}{4n+1}\right).

This is telescoping.

Sn=12 ⁣(1−14n+1)→n→∞1.S_n=\frac12\!\left(1-\frac{1}{4n+1}\right)\xrightarrow[n\to\infty]{}1.

Conclusion: Convergent to 11.

Exercise 3
∑n=1∞19n2+3n−2\sum_{n=1}^{\infty} \frac{1}{9n^2+3n-2}
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Solution

Factor denominator:

9n2+3n−2=(3n−1)(3n+2).9n^2+3n-2=(3n-1)(3n+2).

Telescoping:

1(3n−1)(3n+2)=13 ⁣(13n−1−13n+2).\frac{1}{(3n-1)(3n+2)}=\frac13\!\left(\frac{1}{3n-1}-\frac{1}{3n+2}\right).Sn→16.S_n \to \frac16.

Conclusion: Convergent to 16\frac16.

Exercise 4
∑n=0∞(12n+(−1)n5n)\sum_{n=0}^{\infty}\left(\frac{1}{2^n}+\frac{(-1)^n}{5^n}\right)
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Solution

Sum of two geometric series:

∑12n=2,∑(−1)n5n=11+15=56.\sum \frac{1}{2^n}=2, \qquad \sum \frac{(-1)^n}{5^n}=\frac{1}{1+\frac15}=\frac56.

Conclusion: Convergent to 176\frac{17}{6}.

Exercise 5
∑n=0∞(52n+13n)\sum_{n=0}^{\infty}\left(\frac{5}{2^n}+\frac{1}{3^n}\right)
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Solution

Geometric series:

∑52n=10,∑13n=32.\sum \frac{5}{2^n}=10, \qquad \sum \frac{1}{3^n}=\frac32.

Conclusion: Convergent to 232\frac{23}{2}.

Exercise 6
∑n=1∞n4+13n2+1\sum_{n=1}^{\infty} \frac{\sqrt{n^4+1}}{3n^2+1}
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Solution

Term test:

lim⁡n→∞n4+13n2+1=lim⁡n→∞n23n2=13≠0.\lim_{n\to\infty}\frac{\sqrt{n^4+1}}{3n^2+1} =\lim_{n\to\infty}\frac{n^2}{3n^2}=\frac13\neq0.

Conclusion: Divergent.

Exercise 7
∑n=2∞(1n2−1+(−32)n+1)\sum_{n=2}^{\infty}\left(\frac{1}{n^2-1}+\left(-\frac32\right)^{n+1}\right)
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Solution

Second term is geometric with |r|>1.

Conclusion: Divergent.

Exercise 8
∑n=0∞ln⁡ ⁣(nn+1)\sum_{n=0}^{\infty} \ln\!\left(\frac{n}{n+1}\right)
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Solution

Partial sums:

Sn=ln⁡ ⁣(1n+1)→−∞.S_n=\ln\!\left(\frac{1}{n+1}\right)\to -\infty.

Conclusion: Divergent.

Exercise 9
∑n=1∞ln⁡ ⁣(n2n+1)\sum_{n=1}^{\infty} \ln\!\left(\frac{n}{2n+1}\right)
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Solution
ln⁡ ⁣(n2n+1)→ln⁡ ⁣(12)≠0.\ln\!\left(\frac{n}{2n+1}\right)\to \ln\!\left(\frac12\right)\neq0.

Conclusion: Divergent (Divergence Test).

Exercise 10
∑n=1∞cos⁡(nπ)\sum_{n=1}^{\infty} \cos(n\pi)
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Solution
cos⁡(nπ)=(−1)n↛0.\cos(n\pi)=(-1)^n \nrightarrow 0.

Conclusion: Divergent.

Exercise 11
∑n=1∞2n−13n\sum_{n=1}^{\infty} \frac{2^n-1}{3^n}
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Solution

Split:

∑2n3n−∑13n=∑(23)n−∑(13)n.\sum\frac{2^n}{3^n}-\sum\frac{1}{3^n} =\sum\left(\frac23\right)^n-\sum\left(\frac13\right)^n.

Conclusion: Convergent to 32\frac32.

Exercise 12
∑n=2∞ln⁡ ⁣(1−1n2)\sum_{n=2}^{\infty} \ln\!\left(1-\frac{1}{n^2}\right)
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Solution

Telescoping:

ln⁡ ⁣(1−1n2)=ln⁡ ⁣((n−1)(n+1)n2).\ln\!\left(1-\frac{1}{n^2}\right)=\ln\!\left(\frac{(n-1)(n+1)}{n^2}\right).Sn→ln⁡ ⁣(12).S_n \to \ln\!\left(\frac12\right).

Conclusion: Convergent to ln⁡12\ln\frac12.

Exercise 13
∑n=1∞n+1−nn2+n\sum_{n=1}^{\infty} \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n^2+n}}
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Solution

Simplify:

n+1−nn2+n=1n+1+n.\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n^2+n}} =\frac{1}{\sqrt{n+1}+\sqrt{n}}.

Telescoping ⇒Sn→1\Rightarrow S_n\to1.

Conclusion: Convergent to 11.

Exercise 14
∑n=1∞(1−1n)n\sum_{n=1}^{\infty} \left(1-\frac{1}{n}\right)^n
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Solution
(1−1n)n→e−1≠0.\left(1-\frac{1}{n}\right)^n \to e^{-1}\neq0.

Conclusion: Divergent.

Exercise 15
∑n=1∞2n−13n\sum_{n=1}^{\infty} \frac{2^{n-1}}{3^n}
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Solution

Geometric:

∑2n−13n=13∑(23)n−1.\sum \frac{2^{n-1}}{3^n}=\frac13\sum\left(\frac23\right)^{n-1}.

Conclusion: Convergent to 11.