11.3E: Exercises for Section 11.3

Divergence Test

Consider the sequence for each series in exercises 1 - 14, if the divergence test applies, either state that \(\displaystyle \lim_{n→∞}a_n\) does not exist or find \(\displaystyle \lim_{n→∞}a_n\). If the divergence test does not apply, state why.

1) \(\displaystyle \sum_{n=1}^∞ \dfrac{n}{n+2}\)

Answer

Divergence test applies. Limit of the sequence is 1. Series is divergent.

2) \(\displaystyle \sum_{n=1}^∞\dfrac{n}{5n^2−3}\)

Answer

\(\displaystyle \lim_{n→∞}a_n=0\) or the limit of the sequence is zero. Divergence Test does not apply.

3) \(\displaystyle \sum_{n=1}^∞\dfrac{n}{\sqrt{3n^2+2n+1}}\)

Answer

Limit of the sequence is not zero. Series is divergent by Divergence Test.

4) \(\displaystyle \sum_{n=1}^∞\dfrac{(2n+1)(n−1)}{(n+1)^2}\)

Answer

Limit of the sequence is not zero. Series is divergent by Divergence Test.

5) \(\displaystyle \sum_{n=1}^∞\dfrac{2^n}{3^{n/2}}\)

Answer

The series is geometric series with \(r=2/\sqrt{3}>1\) which is divergent.

6) \(\displaystyle \sum_{n=1}^∞\dfrac{2^n+3^n}{10^{n/2}}\)

Answer

The series is geometric series convergent. Hint: write as two geometric series.

7) \(\displaystyle \sum_{n=1}^∞e^{−2/n}\)

Answer

Series is a geometric series with \(r=1/e^2<1\). It is convergent.

8) \(\displaystyle \sum_{n=1}^∞\cos n\)

Answer

The limit of the sequence does not exist. Series is divergent by Divergence Test.

9) \(\displaystyle \sum_{n=1}^∞\tan n\)

Answer

The limit of the sequence does not exist. Series is divergent by Divergence Test.

10) \(\displaystyle \sum_{n=1}^∞\dfrac{1−\cos^2(1/n)}{\sin^2(2/n)}\)

Answer

Series is divergent.

11) \(\displaystyle \sum_{n=1}^∞\left(1−\dfrac{1}{n}\right)^{2n}\)

Answer

Series is divergent.

12) \(\displaystyle \sum_{n=1}^∞\dfrac{\ln n}{n}\)

Answer

The limit of the sequence is zero. Divergence Test does not apply.

13) \(\displaystyle \sum_{n=1}^∞\dfrac{(\ln n)^2}{\sqrt{n}}\)

Answer

The limit of the sequence is zero. Divergence Test does not apply.

\(p\)-Series & Integral Test

In exercises 14 - 19, state whether the given \( p\)-series converges.

14) \(\displaystyle \sum_{n=1}^∞\frac{1}{\sqrt[6]{n}}\)

Answer

The series is a p-series with p=1/6. It is divergent.

15) \(\displaystyle \sum_{n=1}^∞\frac{1}{n\sqrt{n}}\)

Answer

The series converges, since \( p=3/2>1\).

16) \(\displaystyle \sum_{n=1}^∞\frac{1}{\sqrt[3]{n^2}}\)

Answer

The series is divergent.

17) \(\displaystyle \sum_{n=1}^∞\frac{1}{\sqrt[3]{n^4}}\)

Answer

The series converges, since \( p=4/3>1.\)

18) \(\displaystyle \sum_{n=1}^∞\frac{n^e}{n^π}\)

Answer

The series is divergent.

19) \(\displaystyle \sum_{n=1}^∞\frac{n^π}{n^{2e}}\)

Answer

The series converges, since \( p=2e−π>1.\)

In exercises 20 - 24, use the integral test to determine whether the following sums converge.

20) \(\displaystyle \sum_{n=1}^∞\frac{1}{\sqrt{n+5}}\)

Answer

The series is divergent.

21) \(\displaystyle \sum_{n=1}^∞\frac{1}{\sqrt[3]{n+5}}\)

Answer

The series diverges by the Integral Test since \(\displaystyle ∫^∞_1\frac{dx}{(x+5)^{1/3}}\) can be shown to diverge.

22) \(\displaystyle \sum_{n=2}^∞\frac{1}{n\ln n}\)

Answer

The series is divergent.

23) \(\displaystyle \sum_{n=1}^∞\frac{2n}{1+n^4}\)

Answer

The series converges by the Integral Test since \(\displaystyle ∫^∞_1\frac{2x}{1+x^4}\,dx\) converges.

24) \(\displaystyle \sum_{n=2}^∞\frac{1}{n\ln^2n}\)

Answer

The series is convergent.

Express the sums in exercises 25-26 as \( p\)-series and determine whether each converges.

25) \(\displaystyle \sum_{n=1}^∞2^{−\ln n}\) \(\quad\Big(\)Hint: \( 2^{−\ln n}=\dfrac{1}{n^{\ln 2}}.\Big)\)

Answer

\( 2^{−\ln n}=1/n^{\ln 2}.\) Since \(p=\ln 2<1\), this series diverges by the \( p\)-series test.

26) \(\displaystyle \sum_{n=1}^n2^{−2\ln n}\)

Answer

\( 2^{−2\ln n}=1/n^{2\ln 2}.\) Since \(p = 2\ln 2−1<1\), this series diverges by the \( p\)-series test.

In exercise 27, use the estimate \(\displaystyle R_N≤∫^∞_Nf(t)\,dt\) to find a bound for the remainder \(\displaystyle R_N=\sum_{n=1}^∞a_n−\sum_{n=1}^Na_n\) where \( a_n=f(n)\).

27) \(\displaystyle \sum_{n=1}^{1000}\frac{1}{n^2}\)

Answer

\(\displaystyle R_{1000}≤∫^∞_{1000}\frac{dt}{t^2}=\lim_{b\to ∞}−\frac{1}{t}\bigg|^b_{1000}=\lim_{b\to ∞}\left(−\frac{1}{b}+\frac{1}{1000}\right)=0.001\)

[T] In exercise 28, find the minimum value of \( N\) such that the remainder estimate \(\displaystyle ∫^∞_{N+1}f(x)\,dx<R_N<∫^∞_N f(x)\,dx\) guarantees that \(\displaystyle \sum_{n=1}^Na_n\) estimates \(\displaystyle \sum_{n=1}^∞a_n\), accurate to within the given error.

28) \( a_n=\dfrac{1}{n^2},\) error \( <10^{−4}\)

Answer

\(\displaystyle R_N<∫^∞_N\frac{dx}{x^2}=1/N,\;\text{for}\;N>10^4\)

In exercise 29, find a value of \( N\) such that \( R_N\) is smaller than the desired error. Compute the corresponding sum \(\displaystyle \sum_{n=1}^Na_n\) and compare it to the given estimate of the infinite series.

29) \( a_n=\dfrac{1}{n^4},\) error \(\displaystyle <10^{−4}, \sum_{n=1}^∞\dfrac{1}{n^4}=\frac{π^4}{90}=1.08232...\)

Answer

\(\displaystyle R_N<∫^∞_N\dfrac{dx}{x^4}=4/N^3,\;\text{for}\;N>(4.10^4)^{1/3},\) okay if \( N=35\); \(\displaystyle \sum_{n=1}^{35}\dfrac{1}{n^4}=1.08231….\) Estimate agrees with the sum to four decimal places.