Examples

Ratio test examples

(a) Observe that ${\displaystyle \sum _{n=0}^{\infty }\frac{{10}^n}{n!}}$ has ratio $R_n=\frac{10}{n+1}$ and thus $R_n\to 0=L<1$. Therefore the RaT implies that this series converges.

Simplify the ratio:

$$\begin{array}{c}\frac{{\displaystyle \left(\frac{{10}^{n+1}}{(n+1)!}\right)}}{{\displaystyle \left(\frac{{10}^n}{n!}\right)}}\gg \gg \frac{{10}^{n+1}}{(n+1)!}\cdot \frac{n!}{{10}^n}\\ \\ \gg \gg \frac{10\cdot {10}^n}{(n+1)n!}\cdot \frac{n!}{{10}^n}\gg \gg \frac{10}{n+1}\stackrel{n\to \infty }{⟶}0\end{array}$$

Notice this technique! We frequently use these rules:

$${10}^{n+1}={10}^n\cdot 10,(n+1)!=(n+1)n!$$

(To simplify ratios with exponents and factorials.)

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(b) ${\displaystyle \sum _{n=1}^{\infty }\frac{n^2}{2^n}}$ has ratio $R_n=\frac{{(n+1)}^2}{2^{n+1}}/\frac{n^2}{2^n}$.

Simplify this:

$$\frac{{(n+1)}^2}{2^{n+1}}/\frac{n^2}{2^n}\gg \gg \frac{{(n+1)}^2}{2^{n+1}}\cdot \frac{2^n}{n^2}$$ $$\gg \gg \frac{{(n+1)}^2\cdot 2^n}{n^2\cdot 2\cdot 2^n}\gg \gg \frac{n^2+2n+1}{2n^2}\stackrel{n\to \infty }{⟶}\frac{1}{2}=L$$

So the series converges absolutely by the ratio test.

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(c) Observe that ${\displaystyle \sum _{n=1}^{\infty }{n}^2}$ has ratio $R_n={\displaystyle \frac{n^2+2n+1}{n^2}\to 1}$ as $n\to \infty$.

So the ratio test is inconclusive, even though this series fails the SDT and obviously diverges.

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(d) Observe that ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^2}}$ has ratio $R_n={\displaystyle \frac{n^2}{n^2+2n+1}\to 1}$ as $n\to \infty$.

So the ratio test is inconclusive, even though the series converges as a $p$-series with $p=2>1$.

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(e) More generally, the ratio test is usually inconclusive for rational functions; it is more effective to use LCT with a $p$-series.

Root test examples

(a) Observe that ${\displaystyle \sum _{n=1}^{\infty }{\left(\frac{1}{n}\right)}^n}$ has roots of terms:

$$|a_n{|}^{1/n}={\left({\left(\frac{1}{n}\right)}^n\right)}^{1/n}=\frac{1}{n}\stackrel{n\to \infty }{⟶}0=L$$

Because $L<1$, the RooT shows that the series converges absolutely.

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(b) Observe that ${\displaystyle \sum _{n=1}^{\infty }{(-1)}^n{\left(\frac{n}{2n+1}\right)}^n}$ has roots of terms:

$$\sqrt[n]{|a_n|}=\frac{n}{2n+1}\stackrel{n\to \infty }{⟶}\frac{1}{2}=L$$

Because $L<1$, the RooT shows that the series converges absolutely.

Ratio test versus root test

Determine whether the series ${\displaystyle \sum _{n=1}^{\infty }\frac{n^2{4}^n}{5^{n+2}}}$ converges absolutely or conditionally or diverges.

Solution Before proceeding, rewrite somewhat the general term as ${\displaystyle {\left(\frac{n}{5}\right)}^2\cdot {\left(\frac{4}{5}\right)}^n}$.

Now we solve the problem first using the ratio test. By plugging in $n+1$ we see that

$$a_{n+1}={\left(\frac{n+1}{5}\right)}^2\cdot {\left(\frac{4}{5}\right)}^{n+1}$$

So for the ratio $R_n$ we have:

$$\begin{array}{c}{a}_{n+1}\cdot \frac{1}{a_n}={\left(\frac{n+1}{5}\right)}^2{\left(\frac{4}{5}\right)}^{n+1}\cdot {\left(\frac{5}{n}\right)}^2{\left(\frac{5}{4}\right)}^n\\ \\ \gg \gg \frac{n^2+2n+1}{n^2}\cdot \frac{4}{5}⟶\frac{4}{5}<1\text{ as }n\to \infty \end{array}$$

Therefore the series converges absolutely by the ratio test.

Now solve the problem again using the root test. We have for $\sqrt[n]{|a_n|}$:

$${\left({\left(\frac{n}{5}\right)}^2{\left(\frac{4}{5}\right)}^n\right)}^{1/n}={\left(\frac{n}{5}\right)}^{2/n}\frac{4}{5}$$

To compute the limit as $n\to \infty$ we must use logarithmic limits and L’Hopital’s Rule. So, first take the log:

$$\ln \left({\left(\frac{n}{5}\right)}^{2/n}\frac{4}{5}\right)=\frac{2}{n}\ln \frac{n}{5}+\ln \frac{4}{5}$$

Then for the first term apply L’Hopital’s Rule:

$$\frac{\ln \frac{n}{5}\stackrel{d/dx}{⟶}\frac{1}{n/5}\frac{1}{5}}{n/2\stackrel{d/dx}{⟶}1/2}\gg \gg \frac{1/n}{1/2}\gg \gg \frac{2}{n}⟶0\text{ as }n\to \infty$$

So the first term goes to zero, and the second (constant) term is the value of the limit. So the log limit is $\ln \frac{4}{5}$, and the limit (before taking logs) must be $e^{\ln \frac{4}{5}}$ (inverting the log using $e^x$) and this is $\frac{4}{5}$. Since $\frac{4}{5}<1$, the root test also shows that the series converges absolutely.