W10 - Notes

Ratio test and Root test

Videos

Review Videos

Videos, Math Dr. Bob

• Ratio test: Basics
• Ratio test: Ratio test + DCT
• Root test: Basics
• Root test: for $\sum {(1-1/n^2)}^{n^3}$

Link to original

01 Theory

Theory 1

Ratio Test (RaT)

Applicability: Any series with nonzero terms.

Test Statement:

Suppose that ${\displaystyle \left|\frac{a_{n+1}}{a_n}\right|⟶L}$ as $n\to \infty$.

Then:

$$\begin{array}{cccc}L<1:& & & \sum _{n=1}^{\infty }{a}_n\text{converges absolutely}\\ L>1:& & & \sum _{n=1}^{\infty }{a}_n\text{diverges}\\ & & & \\ L=1\text{or DNE}:& & & \text{test inconclusive}\end{array}$$

Extra - Ratio Test - a deeper look

To understand the ratio test better, first consider this series:

$$\sum _{n=0}^{\infty }\frac{2^n}{n!}=1+\frac{2}{1!}+\frac{2^2}{2!}+\frac{2^3}{3!}+\cdots$$

• The term $\frac{2^3}{3!}$ is created by multiplying the prior term by $\frac{2}{3}$.
• The term $\frac{2^4}{4!}$ is created by multiplying the prior term by $\frac{2}{4}$.
• The term $a_n$ is created by multiplying the prior term by $\frac{2}{n}$.

When $n>3$, the multiplication factor giving the next term is necessarily less than $\frac{2}{3}$. Therefore, when $n>3$, the terms shrink faster than those of a geometric series having $r=\frac{2}{3}$. Therefore this series converges.

Similarly, consider this series:

$$\sum _{n=0}^{\infty }\frac{{10}^n}{n!}=1+\frac{10}{1!}+\frac{{10}^2}{2!}+\frac{{10}^3}{3!}+\cdots$$

Write $R_n=\frac{a_n}{a_{n-1}}$ for the ratio from the prior term $a_{n-1}$ to the current term $a_n$. For this series, $R_n=\frac{10}{n}$.

This ratio falls below $\frac{10}{11}$ when $n>11$, after which the terms necessarily shrink faster than those of a geometric series with $r=\frac{10}{11}$. Therefore this series converges.

The main point of the discussion can be stated like this:

$$R_n\to L<1\text{as}n\to \infty$$

Whenever this is the case, then eventually the ratios are bounded below some $r<1$, and the series terms are smaller than those of a converging geometric series.

Extra - Ratio Test proof

Let us write $R_n=\left|\frac{a_{n+1}}{a_n}\right|$ for the ratio to the next term from term $n$.

Suppose that $R_n\to L$ as $n\to \infty$, and that $L<1$. This means: eventually the ratio of terms is close to $L$; so eventually it is less than $1$.

More specifically, let us define $r=\frac{L+1}{2}$. This is the point halfway between $L$ and $1$. Since $R_n\to L$, we know that eventually $R_n<r$.

Any geometric series with ratio $r$ converges. Set $c=a_N$ for $N$ big enough that $R_N<r$. Then the terms of our series satisfy $|a_{N+n}|\le c{r}^n$, and the series starting from $a_N$ is absolutely convergent by comparison to this geometric series.

(Note that the terms $a_1,\dots ,a_{N-1}$ do not affect convergence.)

Link to original

02 Illustration

Example - Ratio test

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.)

---

(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.

---

(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.

---

(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 $p=2>1$.

---

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

Link to original

03 Theory

Theory 2

Root Test (RooT)

Applicability: Any series.

Test Statement:

Suppose that ${|a_n|}^{1/n}⟶L$ as $n\to \infty$.

Then:

$$\begin{array}{cccc}L<1:& & & \sum _{n=1}^{\infty }{a}_n\text{converges absolutely}\\ L>1:& & & \sum _{n=1}^{\infty }{a}_n\text{diverges}\\ & & & \\ L=1\text{or DNE}:& & & \text{test inconclusive}\end{array}$$

Extra - Root test: explanation

The fact that ${|a_n|}^{1/n}\to L$ and $L<1$ implies that eventually ${|a_n|}^{1/n}<\frac{L+1}{2}$ for all high enough $n$. Set $r=\frac{L+1}{2}$ (the midpoint between $L$ and $1$).

Now, the equation ${|a_n|}^{1/n}<r$ is equivalent to the equation $|a_n|<r^n$.

Therefore, eventually the terms $|a_n|$ are each less than the corresponding terms of this convergent geometric series:

$$\sum _{n=1}^{\infty }{r}^n=1+r+r^2+r^3+\cdots$$

Link to original

04 Illustration

Example - Root test examples

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.

---

(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.

Link to original

Example - Ratio test versus root test

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.

Link to original

Series tests: strategy tips

Videos

Review Videos

Videos, Math Dr. Bob

• Series test round-up: Part I
• Series test round-up: Part II
• Series test round-up: Part III

Videos, Trefor Bazett

• How to choose a series convergence test

Link to original

05 Theory

Theory 1

It can help to associate certain “strategy tips” to find convergence tests based on certain patterns.

Matching powers → Simple Divergence Test

$${\displaystyle \sum _{n=1}^{\infty }\frac{n-1}{2n+1}}$$

Use the SDT because we see the highest power is the same ($=1$) in numerator and denominator.

Rational or Algebraic → Limit Comparison Test

$$\sum _{n=1}^{\infty }\frac{\sqrt{n^3+1}}{3n^3+4n^2+2}$$

Use the LCT because we have a rational or algebraic function (positive terms).

Not rational, not factorials → Integral Test

$$\sum _{n=1}^{\infty }n{e}^{-n^2}$$

Use the IT because we do not have a rational/algebraic function, and we do not see factorials.

Rational, alternating → AST, and LCT or DCT

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

Notice large $n$ behavior is like $n^2/n^4$ or $1/n^2$. This converges. Use the LCT to show absolute convergence. Skip the AST because absolute convergence settles the matter. Lesson: check for absolute BEFORE applying AST, even when alternating!

Factorials → Ratio Test

$$\sum _{n=1}^{\infty }\frac{2^n}{n!}$$

Use the RaT because we see a factorial. (In case of alternating + factorial, use RaT first.)

Recognize geometric → LCT or DCT

$$\sum _{n=1}^{\infty }\frac{1}{2+3^n}$$

Use the LCT or DCT comparing to $\frac{1}{3^n}$ because we see similarity to $\frac{1}{3^n}$ (recognize geometric).

Link to original

Power series: Radius and Interval

Videos

Review Videos

Videos, Math Dr. Bob

• Power series: Interval and Radius of Convergence
• Power series: Interval of Convergence Using Ratio Test
  • Further example
• Power series: Interval of Convergence Using Root Test
• Power series: Finding the Center

Link to original

06 Theory

Theory 1

A power series looks like this:

$$f(x)=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+\cdots$$

Power series are used to build and study functions. They allow a uniform “modeling framework” in which many functions can be described and compared. Power series are also convenient for computers because they provide a way to store and evaluate differentiable functions with numerical (approximate) values.

Small $x$ needed for power series

The most important fact about power series is that they work for small values of $x$.

Many power series diverge for $|x|$ too big; but even when they converge, for big $|x|$ they converge more slowly, and partial sum approximations are less accurate.

The idea of a power series is a modification of the idea of a geometric series in which the common ratio $r$ becomes a variable $x$, and each term has an additional coefficient parameter $a_n$ controlling the relative contribution of different orders.

Link to original

07 Theory

Theory 2

Every power series has a radius of convergence and an interval of convergence.

Radius of convergence

Consider a power series centered at $x=0$:

$$f(x)=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+\cdots$$

Apply the ratio test:

$$\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}{x}^{n+1}}{a_n{x}^n}\right|\gg \gg (\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|)|x|=L$$

Define the radius of convergence $R\in [0,\infty ]$:

$$R=\frac{1}{{\lim }_{n\to \infty }\left|\frac{a_{n+1}}{a_n}\right|}$$

Therefore:

$$\begin{array}{cc}|x|<& R⟹L<1⟹\text{converges}\\ & \\ |x|>& R⟹L>1⟹\text{diverges}\end{array}$$

---

We can build shifted power series for $x$ near some other value $c$. Just replace the variable $x$ with a shifted variable $u=x-c$:

$$\begin{array}{c}{a}_0+a_{1}u+a_2{u}^2+a_3{u}^3+\cdots \\ \\ \gg \gg a_0+a_1(x-c)+a_2{(x-c)}^2+a_3{(x-c)}^3+\cdots \end{array}$$

Now apply the ratio test to determine convergence:

$$\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}|x-c{|}^{n+1}}{a_n|x-c{|}^n}\right|\gg \gg (\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|)|x-c|$$

Define the radius of convergence $R\in [0,\infty ]$:

$$R=\frac{1}{{\lim }_{n\to \infty }\left|\frac{a_{n+1}}{a_n}\right|}$$

In the shifted setting, the radius of convergence limits the *distance from *:

$$\begin{array}{cc}|x-c|<& R⟹\text{converges}\\ & \\ |x-c|>& R⟹\text{diverges}\end{array}$$

---

Method: To calculate the interval of convergence of a power series, follow these steps:

• Observe the center $c$ of the shifted series (or $c=0$ for no shift).
• Compute $R$ using the limit of coefficient ratios.
• Write down the preliminary interval $(c-R,c+R)$.
• Plug each endpoint, $c-R$ and $c-R$, into the original series
  • Check for convergence
• Add in the convergent endpoints. (4 possible scenarios.)

Link to original

08 Illustration

Example - Radius of convergence

Radius of convergence

Find the radius of convergence of the series:

(a) ${\displaystyle \sum _{n=0}^{\infty }\frac{x^n}{2^n}}$ $$ (b) ${\displaystyle \sum _{n=0}^{\infty }\frac{x^{2n}}{(2n)!}}$

Solution

(a) Ratio of terms:

$$\begin{array}{c}\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{{\displaystyle \frac{x^{n+1}}{2^{n+1}}}}{{\displaystyle \frac{x^n}{2^n}}}\right|\\ \\ \gg \gg |\frac{x^{n+1}}{2^{n+1}}\cdot \frac{2^n}{x^n}|\gg \gg \frac{1}{2}|x|=L\end{array}$$

This converges for $L<1$, or $|x|<2$. Therefore $R=2$.

---

(b) This power series skips the odd powers. Apply the ratio test to just the even powers:

$$\begin{array}{c}\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{{\displaystyle \frac{x^{2n+2}}{(2n+2)!}}}{{\displaystyle \frac{x^{2n}}{(2n)!}}}\right|\\ \\ \gg \gg |\frac{x^{2n+2}}{(2n+2)!}\cdot \frac{(2n)!}{x^{2n}}|\\ \\ \gg \gg \frac{1}{(2n+2)(2n+1)}|x^2|\gg \gg R=\infty \end{array}$$ Link to original

Example - Interval of convergence

Interval of convergence

Find the radius and interval of convergence of the following series:

(a) ${\displaystyle \sum _{n=1}^{\infty }\frac{{(x-3)}^n}{n}}$ $$ (b) ${\displaystyle \sum _{n=0}^{\infty }\frac{{(-3)}^n{x}^n}{\sqrt{n+1}}}$

Solution

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

(1) Apply ratio test:

$$\begin{array}{c}|a_{n+1}\cdot a_n^{-1}|=|{\displaystyle \frac{{(x-3)}^{n+1}}{n+1}}\cdot {\displaystyle \frac{n}{{(x-3)}^n}}|\\ \\ \gg \gg \frac{n}{n+1}|x-3|\stackrel{n\to \infty }{⟶}|x-3|=L\end{array}$$

Therefore $R=1$ and thus:

$$\begin{array}{cc}|x-3|<1& ⟹\text{converges}\\ & \\ |x-3|>1& ⟹\text{diverges}\end{array}$$

Preliminary interval: $x\in (\mathrm{2,4})$.

---

(2) Check endpoints:

Check endpoint $x=2$:

$$\begin{array}{c}\sum _{n=1}^{\infty }\frac{{(2-3)}^n}{n}\gg \gg \sum _{n=1}^{\infty }\frac{{(-1)}^n}{n}\\ \\ \gg \gg \text{converges by AST}\end{array}$$

Check endpoint $x=4$:

$$\begin{array}{c}\sum _{n=1}^{\infty }\frac{{(4-3)}^n}{n}\gg \gg \sum _{n=1}^{\infty }\frac{1}{n}\\ \\ \gg \gg \text{diverges as }p\text{-series}\end{array}$$

Final interval of convergence: $x\in [\mathrm{2,4})$

---

(b) ${\displaystyle \sum _{n=0}^{\infty }\frac{{(-3)}^n{x}^n}{\sqrt{n+1}}}$

(1) Apply ratio test:

$$\begin{array}{c}|a_{n+1}\cdot a_n^{-1}|=|\frac{{(-3)}^{n+1}{x}^{n+1}}{\sqrt{n+2}}\cdot \frac{\sqrt{n+1}}{{(-3)}^n{x}^n}|\\ \\ \gg \gg \frac{|(-3){(-3)}^n|}{\sqrt{n+2}}\cdot \frac{\sqrt{n+1}}{|{(-3)}^n|}|x|\\ \\ \gg \gg \frac{3\sqrt{n+1}}{\sqrt{n+2}}|x|\stackrel{n\to \infty }{⟶}3|x|=L\end{array}$$

Therefore:

$$\begin{array}{cc}|x|<\frac{1}{3}& ⟹\text{converges}\\ & \\ |x|>\frac{1}{3}& ⟹\text{diverges}\end{array}$$

Preliminary interval: $x\in (-{\displaystyle \frac{1}{3}},{\displaystyle \frac{1}{3}})$

---

(2) Check endpoints:

Check endpoint $x=-1/3$:

$$\begin{array}{c}\sum _{n=0}^{\infty }\frac{{(-3\cdot (-\frac{1}{3}))}^n}{\sqrt{n+1}}\gg \gg \sum _{n=0}^{\infty }\frac{1}{\sqrt{n+1}}\\ \\ \gg \gg \text{diverges by LCT with }{b}_n=1/\sqrt{n}\end{array}$$

Check endpoint $x=+1/3$:

$$\begin{array}{c}\sum _{n=0}^{\infty }\frac{{(-3\cdot (+\frac{1}{3}))}^n}{\sqrt{n+1}}\gg \gg \sum _{n=0}^{\infty }\frac{{(-1)}^n}{\sqrt{n+1}}\\ \\ \gg \gg \text{converges by AST}\end{array}$$

Final interval of convergence: $x\in (-1/\mathrm{3,1}/3]$

Link to original

Exercise - Radius and interval

Radius and interval for a few series

Find the radius and interval of convergence of the following series:

(a) ${\displaystyle \sum _{n=0}^{\infty }{x}^n}$ $$ (b) ${\displaystyle \sum _{n=0}^{\infty }n!x^n}$

Link to original

Interval of convergence - further examples

Interval of convergence - further examples

Find the interval of convergence of the following series.

(a) ${\displaystyle \sum _{n=0}^{\infty }\frac{n{(x+2)}^n}{3^{n+1}}}$ $$ (b) ${\displaystyle \sum _{n=1}^{\infty }\frac{{(4x+1)}^n}{n}}$

Solution

(a) ${\displaystyle \sum _{n=0}^{\infty }\frac{n{(x+2)}^n}{3^{n+1}}}$

Ratio of terms:

$$\frac{n+1}{3^{n+2}}\cdot \frac{3^{n+1}}{n}|x+2|\gg \gg \frac{n+1}{3n}|x+2|\stackrel{n\to \infty }{⟶}\frac{1}{3}|x+2|=L$$

This converges when $L<1$ or $|x-(-2)|<3$.

Therefore $R=3$ and the preliminary interval is $x\in (-\mathrm{5,1})$.

Check endpoints: ${\displaystyle \sum \frac{n{(-3)}^n}{3^{n+1}}}$ diverges and ${\displaystyle \sum \frac{n{(3)}^n}{3^{n+1}}}$ also diverges.

Final interval is $(-\mathrm{5,1})$.

---

(b) ${\displaystyle \sum _{n=1}^{\infty }\frac{{(4x+1)}^n}{n}}$

Ratio of terms:

$$\begin{array}{c}\left|\frac{a_{n+1}}{a_n}\right|\gg \gg \frac{{\displaystyle \frac{1}{n+1}}}{{\displaystyle \frac{1}{n}}}|4x+1|\\ \\ \gg \gg \frac{n}{n+1}|4x+1|\stackrel{n\to \infty }{⟶}|4x+1|=L\end{array}$$

This converges when $L<1$ or:

$$\begin{array}{cc}|4x+1|<& 1⟺|x+1/4|<1/4⟹\text{converges}\\ & \\ |4x+1|>& 1⟺|x+1/4|>1/4⟹\text{diverges}\end{array}$$

Preliminary interval: $x\in (-1/\mathrm{2,0})$

Check endpoints: ${\displaystyle \sum \frac{{(4\cdot \frac{-1}{2}+1)}^n}{n}}$ converges but ${\displaystyle \sum \frac{1}{n}}$ diverges.

Final interval of convergence: $[-1/\mathrm{2,0})$

Link to original