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Exercise 1: Sequence Convergence and Limits

Theory

Determine whether each sequence converges or diverges. If it converges, find the limit.

Exercises

Exercise 1
{4n−18n+3}\left\{ \frac{4n-1}{8n+3} \right\}
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Solution
lim⁡n→∞4n−18n+3=lim⁡n→∞4−1n8+3n=48=12.\lim_{n\to\infty} \frac{4n-1}{8n+3} = \lim_{n\to\infty} \frac{4-\frac{1}{n}}{8+\frac{3}{n}} = \frac{4}{8} = \frac12.

Convergent to 12\frac12.

Exercise 2
{5(−1)n+1}\left\{ 5(-1)^{n+1} \right\}
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Solution
5(−1)n+1=5,−5,5,−5,…5(-1)^{n+1} = 5,-5,5,-5,\dots

Oscillates ⇒\Rightarrow divergent.

Exercise 3
{n3−12n}\left\{ \frac{n^3-1}{2n} \right\}
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Solution
lim⁡n→∞n3−12n=lim⁡n→∞n22=∞.\lim_{n\to\infty} \frac{n^3-1}{2n} = \lim_{n\to\infty} \frac{n^2}{2} = \infty.

Divergent.

Exercise 4
{2n+1n}\left\{ \frac{\sqrt{2n+1}}{n} \right\}
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Solution
lim⁡n→∞2n+1n=lim⁡n→∞n2+1nn=lim⁡n→∞2+1nn=0.\lim_{n\to\infty} \frac{\sqrt{2n+1}}{n} = \lim_{n\to\infty} \frac{\sqrt{n}\sqrt{2+\frac1n}}{n} = \lim_{n\to\infty} \frac{\sqrt{2+\frac1n}}{\sqrt{n}} = 0.

Convergent to 00.

Exercise 5
{ln⁡nn}\left\{ \frac{\ln n}{n} \right\}
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Solution
lim⁡n→∞ln⁡nn=0.\lim_{n\to\infty} \frac{\ln n}{n} = 0.

Convergent to 00.

Exercise 6
{6−2−n3+4−n}\left\{ \frac{6-2^{-n}}{3+4^{-n}} \right\}
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Solution
lim⁡n→∞6−2−n3+4−n=63=2.\lim_{n\to\infty} \frac{6-2^{-n}}{3+4^{-n}} = \frac{6}{3} = 2.

Convergent to 22.

Exercise 7
{n2n+1}\left\{ n^{\frac{2}{n+1}} \right\}
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Solution
lim⁡n→∞n2n+1=elim⁡n→∞2ln⁡nn+1=e0=1.\lim_{n\to\infty} n^{\frac{2}{n+1}} = e^{\lim_{n\to\infty} \frac{2\ln n}{n+1}} = e^0 = 1.

Convergent to 11.

Exercise 8
{ln⁡ ⁣(4n+15n−1)}\left\{ \ln\!\left(\frac{4n+1}{5n-1}\right) \right\}
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Solution
lim⁡n→∞ln⁡ ⁣(4n+15n−1)=ln⁡ ⁣(45).\lim_{n\to\infty} \ln\!\left(\frac{4n+1}{5n-1}\right) = \ln\!\left(\frac45\right).

Convergent to ln⁡(4/5)\ln(4/5).

Exercise 9
{sin⁡2n4n}\left\{ \frac{\sin^2 n}{4^n} \right\}
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Solution
0≤sin⁡2n4n≤14n.0 \le \frac{\sin^2 n}{4^n} \le \frac{1}{4^n}.

Since lim⁡n→∞14n=0\lim_{n\to\infty} \frac{1}{4^n}=0, by the Sandwich Theorem,

lim⁡n→∞sin⁡2n4n=0.\lim_{n\to\infty} \frac{\sin^2 n}{4^n} = 0.
Exercise 10
{(−1)n5n3n3+1}\left\{ (-1)^n \frac{5n^3}{n^3+1} \right\}
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Solution
5n3n3+1→5but sign alternates.\frac{5n^3}{n^3+1} \to 5 \quad \text{but sign alternates}.

Oscillates ⇒\Rightarrow divergent.

Exercise 11
{nsin⁡πn}\left\{ n\sin\frac{\pi}{n} \right\}
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Solution
lim⁡n→∞nsin⁡πn=π.\lim_{n\to\infty} n\sin\frac{\pi}{n} = \pi.

Convergent to π\pi.

Exercise 12
{(−1)n+1n2}\left\{ \frac{(-1)^{n+1}}{n^2} \right\}
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Solution
∣(−1)n+1n2∣≤1n2→0.\left| \frac{(-1)^{n+1}}{n^2} \right| \le \frac{1}{n^2} \to 0.

Convergent to 00.

Exercise 13
{en4n}\left\{ \frac{e^n}{4^n} \right\}
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Solution
lim⁡n→∞(e4)n=0.\lim_{n\to\infty} \left(\frac{e}{4}\right)^n = 0.

Convergent to 00.

Exercise 14
{n2+3n−n}\left\{ \sqrt{n^2+3n}-n \right\}
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Solution
n2+3n−n=3nn2+3n+n.\sqrt{n^2+3n}-n = \frac{3n}{\sqrt{n^2+3n}+n}.lim⁡n→∞3nn2+3n+n=32.\lim_{n\to\infty} \frac{3n}{\sqrt{n^2+3n}+n} = \frac{3}{2}.

Convergent to 32\frac32.

Exercise 15
{(n+3n+1)n}\left\{ \left(\frac{n+3}{n+1}\right)^n \right\}
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Solution
(n+3n+1)n=(1+2n+1)n.\left(\frac{n+3}{n+1}\right)^n = \left(1+\frac{2}{n+1}\right)^n.lim⁡n→∞(1+2n+1)n=e2.\lim_{n\to\infty} \left(1+\frac{2}{n+1}\right)^n = e^2.

Convergent to e2e^2.