Limit Comparison Test

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Calculus

Cegep/2

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452 words

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3 minutes

Test to determine whether a series diverges by considering a simplified general term with same convergence / divergence

Let $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ and $\sum _{n=1}^{\mathrm{\infty }}{b}_n$ be two series with only positive terms.
If $\underset{n\to \mathrm{\infty }}{lim}\frac{a_n}{b_n}$ exists and is non zero, then $\sum a\sim \sum b$.

Examples

Determine if the following series are convergent:

$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^3+1}$

$$\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^3+1}& ⟹a_n=\frac{1}{n^3+1}\\ \sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^3}& ⟹b_n=\frac{1}{n^3}\\ \underset{n\to \mathrm{\infty }}{lim}\frac{a_n}{b_n}& =\underset{n\to \mathrm{\infty }}{lim}\frac{\frac{1}{n^3+1}}{\frac{1}{n^3}}\\ & =\underset{n\to \mathrm{\infty }}{lim}\frac{n^3}{n^3+1}\\ & =1\ne 0\end{array}$$

Therefore, $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^3+1}$ is also convergent by LCT.

$\sum _{n=1}^{\mathrm{\infty }}\cos \frac{1}{\sqrt{n}}$

$$\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}\cos \frac{1}{\sqrt{n}}& ⟹a_n=\cos \frac{1}{\sqrt{n}}\\ \sum _{n=1}^{\mathrm{\infty }}1& ⟹b_n=1\\ \underset{n\to \mathrm{\infty }}{lim}\frac{a_n}{b_n}& =\underset{n\to \mathrm{\infty }}{lim}\frac{\cos \frac{1}{\sqrt{n}}}{1}\\ & =\cos \underset{n\to \mathrm{\infty }}{lim}\frac{1}{\sqrt{n}}\\ & =\cos 0\\ & =1\ne 0\end{array}$$

Therefore, $\sum _{n=1}^{\mathrm{\infty }}\cos \frac{1}{\sqrt{n}}$ is also divergent by LCT.

$\sum _{n=1}^{\mathrm{\infty }}\frac{3n+4^n}{5^n+5}$

$$\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}\frac{3n+4^n}{5^n+5}& ⟹a_n=\frac{3n+4^n}{5^n+5}\\ \sum _{n=1}^{\mathrm{\infty }}\frac{4^n}{5^n}& ⟹b_n=\frac{4^n}{5^n}\\ |r|& =\frac{4}{5}<1\therefore \text{convergent geometric series}\\ \\ \underset{n\to \mathrm{\infty }}{lim}\left(\frac{3n+4^n}{5^n+5}\right)\left(\frac{5^n}{4^n}\right)& =\underset{n\to \mathrm{\infty }}{lim}\frac{\frac{3n}{4^n}+1}{1+\frac{5}{5^n}}\\ & =1\ne 0\end{array}$$

Therefore, $\sum a_n$ is also convergent by LCT.

$\sum _{n=1}^{\mathrm{\infty }}(1+\frac{1}{n})e^{-n}$

$$\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}(1+\frac{1}{n})e^{-n}& ⟹a_n=(1+\frac{1}{n})e^{-n}\\ \sum _{n=1}^{\mathrm{\infty }}{e}^{-1}& ⟹b_n=e^{-1}\\ |r|& =\frac{1}{e}<1\therefore \text{convergent geometric series}\\ \\ \underset{n\to \mathrm{\infty }}{lim}\frac{(1+\frac{1}{n})e^{-n}}{e^{-n}}& =\underset{n\to \mathrm{\infty }}{lim}(1+\frac{1}{n})\\ & =1\ne 0\end{array}$$

Therefore, $\sum a_n$ is also convergent by LCT.

Contributors

Ajitani Hifumi

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