About

Limits and Convergence

Theory

A real number LL is the limit of a sequence {an}\{a_n\} if ana_n becomes arbitrarily close to LL as n→∞n \to \infty.
If the limit exists, the sequence is convergent; otherwise, it is divergent.

Beginner Computation Pattern

  1. Simplify using dominant terms
  2. Take the limit as n→∞n \to \infty
  3. State convergence or divergence

Exercises

Exercise 1

Check whether

{1−6n4n4+8n3}\left\{ \frac{1-6n^4}{n^4+8n^3} \right\}

converges or diverges.

Show solution ↓
Solution

Simplify:

1−6n4n4+8n3=1/n4−61+8/n.\frac{1-6n^4}{n^4+8n^3} = \frac{1/n^4 - 6}{1 + 8/n}.

Take the limit:

lim⁡n→∞1/n4−61+8/n=−6.\lim_{n\to\infty} \frac{1/n^4 - 6}{1 + 8/n} = -6.

Conclusion: The sequence converges to −6-6.

Exercise 2

Check whether

{2n2−2n+1n−1}\left\{ \frac{2n^2-2n+1}{n-1} \right\}

converges or diverges.

Show solution ↓
Solution

Simplify: The degree of the numerator exceeds the denominator.

Take the limit:

lim⁡n→∞2n2−2n+1n−1=∞.\lim_{n\to\infty} \frac{2n^2-2n+1}{n-1} = \infty.

Conclusion: The sequence diverges.

Exercise 3

Check whether

{21000+2n−1+3n−22n+3n+5}\left\{ \frac{2^{1000}+2^{n-1}+3^{n-2}}{2^n+3^n+5} \right\}

converges or diverges.

Show solution ↓
Solution

Simplify: Dominant terms are 3n−23^{n-2} and 3n3^n.

21000+2n−1+3n−22n+3n+5∼3n−23n.\frac{2^{1000}+2^{n-1}+3^{n-2}}{2^n+3^n+5} \sim \frac{3^{n-2}}{3^n}.

Take the limit:

lim⁡n→∞3n−23n=19.\lim_{n\to\infty} \frac{3^{n-2}}{3^n} = \frac{1}{9}.

Conclusion: The sequence converges to 19\frac{1}{9}.

Exercise 4

Check whether

{n−n2−n}\left\{ n-\sqrt{n^2-n} \right\}

converges or diverges.

Show solution ↓
Solution

Simplify: Multiply by the conjugate.

n−n2−n=nn+n2−n.n-\sqrt{n^2-n} = \frac{n}{n+\sqrt{n^2-n}}.

Take the limit:

lim⁡n→∞nn+n2−n=12.\lim_{n\to\infty} \frac{n}{n+\sqrt{n^2-n}} = \frac12.

Conclusion: The sequence converges to 12\frac12.

Exercise 5

Check whether

{ln⁡n−ln⁡(2n3+1)}\left\{ \ln n - \ln(2n^3+1) \right\}

converges or diverges.

Show solution ↓
Solution

Simplify:

ln⁡n−ln⁡(2n3+1)=ln⁡ ⁣(n2n3+1).\ln n - \ln(2n^3+1) = \ln\!\left(\frac{n}{2n^3+1}\right).

Take the limit:

lim⁡n→∞ln⁡ ⁣(n2n3+1)=−∞.\lim_{n\to\infty} \ln\!\left(\frac{n}{2n^3+1}\right) = -\infty.

Conclusion: The sequence diverges.

Exercise 6

Find the limit of the sequence

{(n+1n−1)n}.\left\{ \left(\frac{n+1}{n-1}\right)^n \right\}.
Show solution ↓
Solution

Simplify:

(n+1n−1)n=(1+2n−1)n.\left(\frac{n+1}{n-1}\right)^n = \left(1+\frac{2}{n-1}\right)^n.

Take the limit:

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

Conclusion: The sequence converges to e2e^2.

Exercise 7

Show that

{sin⁡nn}\left\{ \frac{\sin n}{n} \right\}

converges.

Show solution ↓
Solution

Simplify:

−1n≤sin⁡nn≤1n.-\frac{1}{n} \le \frac{\sin n}{n} \le \frac{1}{n}.

Take the limit:

lim⁡n→∞(−1n)=lim⁡n→∞1n=0.\lim_{n\to\infty} \left(-\frac{1}{n}\right) = \lim_{n\to\infty} \frac{1}{n} = 0.

Conclusion: By the Sandwich Theorem, the sequence converges to 00.

Exercise 8

Does

{sin⁡ ⁣(nπ+22n)}\left\{ \sin\!\left(\frac{n\pi+2}{2n}\right) \right\}

converge or diverge?

Show solution ↓
Solution

Simplify:

nπ+22n→π2.\frac{n\pi+2}{2n} \to \frac{\pi}{2}.

Take the limit:

lim⁡n→∞sin⁡ ⁣(nπ+22n)=1.\lim_{n\to\infty} \sin\!\left(\frac{n\pi+2}{2n}\right) = 1.

Conclusion: The sequence converges to 11.