Examples

p-series examples

By finding $p$ and applying the $p$-series convergence properties:

We see that ${\sum }_{n=1}^{\infty }\frac{1}{n^{1.1}}$ converges: $p=1.1$ so $p>1$

But ${\sum }_{n=1}^{\infty }\frac{1}{\sqrt{n}}$ diverges: $p=1/2$ so $p\le 1$

Integral test - pushing the envelope of convergence

Does ${\displaystyle \sum _{n=2}^{\infty }\frac{1}{n\ln n}}$ converge?

Does ${\displaystyle \sum _{n=2}^{\infty }\frac{1}{n{(\ln n)}^2}}$ converge?

Notice that $\ln n$ grows very slowly with $n$, so $\frac{1}{n\ln n}$ is just a little smaller than $\frac{1}{n}$ for large $n$, and similarly $\frac{1}{n{(\ln n)}^2}$ is just a little smaller still.

Solution

(1) The two series lead to the two functions $f(x)=\frac{1}{x\ln x}$ and $g(x)=\frac{1}{x{(\ln x)}^2}$.

Check applicability.

Clearly $f(x)$ and $g(x)$ are both continuous, positive, decreasing functions on $x\in [2,\infty )$.

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(2) Apply the integral test to $f(x)$.

Integrate $f(x)$:

$$\begin{array}{cc}{\int }_2^{\infty }\frac{1}{x\ln x}dx& \gg \gg {\int }_{u=\ln 2}^{\infty }\frac{1}{u}du\\ & \\ & \gg \gg \underset{R\to \infty }{\lim }\ln u{|}_{\ln 1}^R\gg \gg \infty \end{array}$$

Conclude: ${\sum }_{n=2}^{\infty }\frac{1}{n\ln n}$ diverges.

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(3) Apply the integral test to $g(x)$.

Integrate $g(x)$:

$$\begin{array}{cc}{\int }_2^{\infty }\frac{1}{x{(\ln x)}^2}dx& \gg \gg {\int }_{u=\ln 2}^{\infty }\frac{1}{u^2}du\\ & \\ & \gg \gg \underset{R\to \infty }{\lim }-u^{-1}{|}_{\ln 2}^R\gg \gg \frac{1}{\ln 2}\end{array}$$

Conclude: ${\sum }_{n=2}^{\infty }\frac{1}{n{(\ln n)}^2}$ converges.

Direct comparison test: rational functions

(a) ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{\sqrt{n}{3}^n}}$

Choose: $a_n=\frac{1}{\sqrt{n}{3}^n}$ and $b_n=\frac{1}{3^n}$

Check: $\frac{1}{\sqrt{n}{3}^n}\le \frac{1}{3^n}$

Observe: $\sum \frac{1}{3^n}$ is a convergent geometric series

Therefore: converges by the DCT.

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(b) ${\displaystyle \sum _{n=1}^{\infty }\frac{{\cos }^{2}n}{n^3}}$

Choose: $a_n=\frac{{\cos }^{2}n}{n^3}$ and $b_n=\frac{1}{n^3}$.

Check: $\frac{{\cos }^{2}n}{n^3}\le \frac{1}{n^3}$

Observe: $\sum \frac{1}{n^3}$ is a convergent $p$-series

Therefore: converges by the DCT.

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(c) ${\displaystyle \sum _{n=1}^{\infty }\frac{n}{n^3+1}}$

Choose: $a_n=\frac{n}{n^3+1}$ and $b_n=\frac{1}{n^2}$

Check: $\frac{n}{n^3+1}\le \frac{1}{n^2}$ (notice that $\frac{n}{n^3+1}\le \frac{n}{n^3}$)

Observe: $\sum \frac{1}{n^2}$ is a convergent $p$-series

Therefore: converges by the DCT.

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(d) ${\displaystyle \sum _{n=2}^{\infty }\frac{1}{n-1}}$

Choose: $a_n=\frac{1}{n}$ and $b_n=\frac{1}{n-1}$

Check: $\frac{1}{n}\le \frac{1}{n-1}$

Observe: $\sum \frac{1}{n}$ is a divergent $p$-series

Therefore: diverges by the DCT.

Limit comparison test examples

(a) ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{2^n-1}}$

Choose: $a_n=\frac{1}{2^n-1}$ and $b_n=\frac{1}{2^n}$.

Compare in the limit:

$$\underset{n\to \infty }{\lim }\frac{a_n}{b_n}\gg \gg \underset{n\to \infty }{\lim }\frac{2^n}{2^n-1}\gg \gg 1=:L$$

Observe: $\sum \frac{1}{2^n}$ is a convergent geometric series

Therefore: converges by the LCT.

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(b) ${\displaystyle \sum _{n=1}^{\infty }\frac{2n^2+3n}{\sqrt{5+n^5}}}$

Choose: $a_n=\frac{2n^2+3n}{\sqrt{5+n^5}}$, $b_n=\frac{1}{n^{1/2}}$

Compare in the limit:

$$\underset{n\to \infty }{\lim }\frac{a_n}{b_n}\gg \gg \underset{n\to \infty }{\lim }\frac{(2n^2+3n)\sqrt{n}}{\sqrt{5+n^5}}$$ $$\frac{(2n^2+3n)\sqrt{n}}{\sqrt{5+n^5}}\stackrel{n\to \infty }{⟶}\frac{2n^{5/2}}{n^{5/2}}\to 2=:L$$

Observe: $\sum \frac{1}{n^{1/2}}$ is a divergent $p$-series

Therefore: diverges by the LCT.

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(c) ${\displaystyle \sum _{n=2}^{\infty }\frac{n^2}{n^4-n-1}}$

Choose: $a_n=\frac{n^2}{n^4-n-1}$ and $b_n=\frac{1}{n^2}$

Compare in the limit:

$$\underset{n\to \infty }{\lim }\frac{a_n}{b_n}\gg \gg \underset{n\to \infty }{\lim }\frac{n^4}{n^4-n-1}\gg \gg 1=:L$$

Observe: ${\sum }_{n=2}^{\infty }\frac{1}{n^2}$ is a converging $p$-series

Therefore: converges by the LCT.