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Exercises: Comparison Test

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Exercises

Exercise 1
∑n=1∞sin⁡2n2n\sum_{n=1}^{\infty} \frac{\sin^{2} n}{2^{n}}
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Solution

Since 0≤sin⁡2n≤10 \le \sin^2 n \le 1,

0≤sin⁡2n2n≤12n,0 \le \frac{\sin^2 n}{2^n} \le \frac{1}{2^n},

and ∑12n\sum \frac{1}{2^n} converges. Hence, the series converges.

Exercise 2
∑n=1∞1−nn2n\sum_{n=1}^{\infty} \frac{1-n}{n2^{n}}
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Solution
∣1−nn2n∣≤12n.\left|\frac{1-n}{n2^n}\right| \le \frac{1}{2^n}.

Since ∑12n\sum \frac{1}{2^n} converges, the series converges.

Exercise 3
∑n=1∞13n−1+1\sum_{n=1}^{\infty} \frac{1}{3^{n-1}+1}
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Solution
13n−1+1≤13n−1.\frac{1}{3^{n-1}+1} \le \frac{1}{3^{n-1}}.

Geometric series ∑13n−1\sum \frac{1}{3^{n-1}} converges, so the series converges.

Exercise 4
∑n=1∞sin⁡1n\sum_{n=1}^{\infty} \sin \frac{1}{n}
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Solution

For small xx, sin⁡x∼x\sin x \sim x, so

sin⁡1n∼1n.\sin\frac{1}{n} \sim \frac{1}{n}.

Since ∑1n\sum \frac{1}{n} diverges, the series diverges.

Exercise 5
∑n=1∞11+2+3+⋯+n\sum_{n=1}^{\infty} \frac{1}{1+2+3+\dots+n}
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Solution
1+2+⋯+n=n(n+1)2,11+2+⋯+n≤2n2.1+2+\cdots+n = \frac{n(n+1)}{2}, \quad \frac{1}{1+2+\cdots+n} \le \frac{2}{n^2}.

Since ∑1n2\sum \frac{1}{n^2} converges, the series converges.

Exercise 6
∑n=1∞n−1n3+1\sum_{n=1}^{\infty} \frac{n-1}{n^{3}+1}
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Solution
n−1n3+1≤nn3=1n2.\frac{n-1}{n^3+1} \le \frac{n}{n^3}=\frac{1}{n^2}.

Since ∑1n2\sum \frac{1}{n^2} converges, the series converges.

Exercise 7
∑n=2∞nn+1\sum_{n=2}^{\infty} \frac{\sqrt{n}}{n+1}
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Solution
nn+1∼1n.\frac{\sqrt{n}}{n+1} \sim \frac{1}{\sqrt{n}}.

Since ∑1n\sum \frac{1}{\sqrt{n}} diverges, the series diverges.

Exercise 8
∑n=1∞1n3+1\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^{3}+1}}
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Solution
1n3+1≤1n3/2.\frac{1}{\sqrt{n^3+1}} \le \frac{1}{n^{3/2}}.

Since ∑1n3/2\sum \frac{1}{n^{3/2}} converges, the series converges.

Exercise 9
∑n=1∞3+cos⁡n3n\sum_{n=1}^{\infty} \frac{3+\cos n}{3^{n}}
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Solution
0≤3+cos⁡n≤4,3+cos⁡n3n≤43n.0 \le 3+\cos n \le 4, \quad \frac{3+\cos n}{3^n} \le \frac{4}{3^n}.

Geometric series converges, so the series converges.

Exercise 10
∑n=1∞sin⁡2nnn\sum_{n=1}^{\infty} \frac{\sin^{2} n}{n\sqrt{n}}
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Solution
0≤sin⁡2nnn≤1n3/2.0 \le \frac{\sin^2 n}{n\sqrt{n}} \le \frac{1}{n^{3/2}}.

Since ∑1n3/2\sum \frac{1}{n^{3/2}} converges, the series converges.

Exercise 11
∑n=1∞2n1+3n\sum_{n=1}^{\infty} \frac{2^{n}}{1+3^{n}}
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Solution
2n1+3n≤(23)n.\frac{2^n}{1+3^n} \le \left(\frac{2}{3}\right)^n.

Geometric series converges, so the series converges.

Exercise 12
∑n=1∞arctan⁡nn3\sum_{n=1}^{\infty} \frac{\arctan n}{n^{3}}
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Solution

Since 0<\arctan n \le \frac{\pi}{2},

arctan⁡nn3≤π2n3.\frac{\arctan n}{n^3} \le \frac{\pi}{2n^3}.

Since ∑1n3\sum \frac{1}{n^3} converges, the series converges.

Exercise 13
∑n=1∞1+2n1+3n\sum_{n=1}^{\infty} \frac{1+2^{n}}{1+3^{n}}
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Solution
1+2n1+3n≤2n3n=(23)n.\frac{1+2^n}{1+3^n} \le \frac{2^n}{3^n}=\left(\frac{2}{3}\right)^n.

Geometric series converges, so the series converges.

Exercise 14
∑n=1∞n(n+1)2n\sum_{n=1}^{\infty} \frac{n}{(n+1)2^{n}}
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Solution
n(n+1)2n≤12n.\frac{n}{(n+1)2^n} \le \frac{1}{2^n}.

Geometric series converges, so the series converges.

Exercise 15
∑n=1∞n+1n2n\sum_{n=1}^{\infty} \frac{n+1}{n2^{n}}
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Solution
n+1n2n≤22n.\frac{n+1}{n2^n} \le \frac{2}{2^n}.

Geometric series converges, so the series converges.

Exercise 16
∑n=1∞n!n2\sum_{n=1}^{\infty} \frac{n!}{n^{2}}
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Solution

Since n!n! grows faster than any power of nn, the terms do not tend to zero.
Hence, the series diverges.

Exercise 17
∑n=2∞1ln⁡n\sum_{n=2}^{\infty} \frac{1}{\ln n}
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Solution
1ln⁡n≥1nfor large n.\frac{1}{\ln n} \ge \frac{1}{n} \quad \text{for large } n.

Since ∑1n\sum \frac{1}{n} diverges, the series diverges.

Exercise 18
∑n=1∞1nn\sum_{n=1}^{\infty} \frac{1}{n^{n}}
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Solution
1nn≤12nfor n≥2.\frac{1}{n^n} \le \frac{1}{2^n} \quad \text{for } n\ge2.

Geometric series converges, so the series converges.

Exercise 19
∑n=1∞2n−13n+5n\sum_{n=1}^{\infty} \frac{2^{n}-1}{3^{n}+5^{n}}
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Solution
2n−13n+5n≤2n3n=(23)n.\frac{2^n-1}{3^n+5^n} \le \frac{2^n}{3^n}=\left(\frac{2}{3}\right)^n.

Geometric series converges, so the series converges.

Exercise 20
∑n=1∞3sin⁡2nn!\sum_{n=1}^{\infty} \frac{3\sin^{2} n}{n!}
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Solution
0≤3sin⁡2nn!≤3n!.0 \le \frac{3\sin^2 n}{n!} \le \frac{3}{n!}.

Since ∑1n!\sum \frac{1}{n!} converges, the series converges.