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Exercises: Integral Test

Theory

Determine whether the following series converge or diverge.

Exercises

Exercise 1
∑n=1∞nn+1\sum_{n=1}^{\infty} \frac{n}{n+1}
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Solution

Let f(x)=xx+1f(x)=\frac{x}{x+1}.

∫1∞xx+1 dx=∫1∞(1−1x+1)dx=∞.\int_1^\infty \frac{x}{x+1}\,dx = \int_1^\infty \left(1-\frac{1}{x+1}\right)dx = \infty.

Conclusion: The series diverges.

Exercise 2
∑n=1∞11+4n2\sum_{n=1}^{\infty} \frac{1}{1+4n^{2}}
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Solution

Let f(x)=11+4x2f(x)=\frac{1}{1+4x^2}.

\int_1^\infty \frac{1}{1+4x^2}\,dx < \int_0^\infty \frac{1}{1+4x^2}\,dx = \frac{\pi}{4}.

Conclusion: The series converges.

Exercise 3
∑n=1∞1n+5\sum_{n=1}^{\infty} \frac{1}{\sqrt{n+5}}
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Solution

Let f(x)=1x+5f(x)=\frac{1}{\sqrt{x+5}}.

∫1∞1x+5 dx=∞.\int_1^\infty \frac{1}{\sqrt{x+5}}\,dx = \infty.

Conclusion: The series diverges.

Exercise 4
∑n=1∞nln⁡(n+1)\sum_{n=1}^{\infty} \frac{n}{\ln(n+1)}
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Solution

Let f(x)=xln⁡(x+1)f(x)=\frac{x}{\ln(x+1)}. Since f(x)f(x) grows without bound,

∫1∞xln⁡(x+1) dx=∞.\int_1^\infty \frac{x}{\ln(x+1)}\,dx = \infty.

Conclusion: The series diverges.

Exercise 5
∑n=1∞ln⁡nn2\sum_{n=1}^{\infty} \frac{\ln n}{n^{2}}
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Solution

Let f(x)=ln⁡xx2f(x)=\frac{\ln x}{x^2}.

\int_1^\infty \frac{\ln x}{x^2}\,dx < \infty.

Conclusion: The series converges.

Exercise 6
∑n=1∞en1+e2n\sum_{n=1}^{\infty} \frac{e^{n}}{1+e^{2n}}
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Solution

For large nn,

en1+e2n∼e−n.\frac{e^n}{1+e^{2n}} \sim e^{-n}.

Since

\int_1^\infty e^{-x}\,dx < \infty,

Conclusion: The series converges.

Exercise 7
∑n=1∞8tan⁡−1nn2+1\sum_{n=1}^{\infty} \frac{8\tan^{-1} n}{n^{2}+1}
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Solution

Since tan⁡−1n≤π2\tan^{-1}n \le \frac{\pi}{2},

8tan⁡−1nn2+1≤4πn2.\frac{8\tan^{-1} n}{n^2+1} \le \frac{4\pi}{n^2}.

Also,

\int_1^\infty \frac{1}{x^2}\,dx < \infty.

Conclusion: The series converges.

Exercise 8
∑n=2∞1nln⁡n\sum_{n=2}^{\infty} \frac{1}{n\ln n}
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Solution

Let f(x)=1xln⁡xf(x)=\frac{1}{x\ln x}.

∫2∞1xln⁡x dx=ln⁡(ln⁡x)∣2∞=∞.\int_2^\infty \frac{1}{x\ln x}\,dx = \ln(\ln x)\Big|_2^\infty = \infty.

Conclusion: The series diverges.

Exercise 9
∑n=3∞1nln⁡nln⁡2n−1\sum_{n=3}^{\infty} \frac{1}{n\ln n\sqrt{\ln^2 n-1}}
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Solution

Let u=ln⁡xu=\ln x.

∫1xln⁡xln⁡2x−1 dx=∫1uu2−1 du.\int \frac{1}{x\ln x\sqrt{\ln^2 x-1}}\,dx = \int \frac{1}{u\sqrt{u^2-1}}\,du.

This integral converges.

Conclusion: The series converges.

Exercise 10
∑n=1∞1(n+1)ln⁡2(n+1)\sum_{n=1}^{\infty} \frac{1}{(n+1)\ln^2(n+1)}
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Solution

Let f(x)=1xln⁡2xf(x)=\frac{1}{x\ln^2 x}.

\int_2^\infty \frac{1}{x\ln^2 x}\,dx < \infty.

Conclusion: The series converges.

Exercise 11

Show that

∑n=2∞1nln⁡pn\sum_{n=2}^{\infty} \frac{1}{n\ln^p n}

converges if p>1 and diverges if p≤1p\le1.

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Solution

Let f(x)=1xln⁡pxf(x)=\frac{1}{x\ln^p x}.

∫1xln⁡px dx={finite,pgt;1,∞,p≤1.\int \frac{1}{x\ln^p x}\,dx = \begin{cases} \text{finite}, & p>1,\\ \infty, & p\le1. \end{cases}

Conclusion: The series converges for p>1 and diverges for p≤1p\le1.

Exercise 12

Show that

∑n=3∞1nln⁡n(ln⁡p(ln⁡n))\sum_{n=3}^{\infty} \frac{1}{n\ln n(\ln^p(\ln n))}

converges if p>1 and diverges if p≤1p\le1.

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Solution

Let u=ln⁡(ln⁡x)u=\ln(\ln x).

∫1xln⁡x(ln⁡p(ln⁡x)) dx=∫1up du.\int \frac{1}{x\ln x(\ln^p(\ln x))}\,dx = \int \frac{1}{u^p}\,du.

This integral converges if p>1 and diverges if p≤1p\le1.

Conclusion: The series converges for p>1 and diverges for p≤1p\le1.