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.)

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:

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