**d’Alembert-Cauchy Ratio Test**

The following d’Alembert-Cauchy ratio test is one of the easiest to apply and is widely used.

*Theorem* (d’Alembert-Cauchy Ratio Test). Suppose that $\sum_{n=1}^\infty a_n$ is a series with positive terms.

- If $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}<1$ then $\sum_{n=1}^\infty a_n$ converges.
- If $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}>1$ then $\sum_{n=1}^\infty a_n$ diverges.
- If $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=1$ then $\sum_{n=1}^\infty a_n$ then the convergence is indeterminant, i.e., the ratio test provides no information regarding the convergence of the series $\sum_{n=1}^\infty a_n$.

*Example*. Test $\sum_{n=1}^\infty\frac{n}{2^n}$ for convergence.

*Solution*. \begin{align*}\lim_{n\to\infty}\frac{a_{n+1}}{a_n}&=\lim_{n\to\infty}\frac{\frac{n+1}{2^{n+1}}}{\frac{n}{2^n}}\\&=\lim_{n\to\infty}\frac{n+1}{2n}\\&=\frac{1}{2}<1\end{align*} Hence by the Ratio test the series converges.

*Example*. Test the convergence of the series $\sum_{n=1}^\infty\frac{n^n}{n!}$.

*Solution*.

\begin{align*}

\lim_{n\to\infty}\frac{a_{n+1}}{a_n}&=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n\\

&=e>1.

\end{align*}

Hence, the series diverges.

*Remark*. There is an easier way to show the divergence of the series $\sum_{n=1}^\infty\frac{n^n}{n!}$.

Note that

$$a_n=\frac{n^n}{n!}=\frac{n\cdot n\cdot n\cdots n}{1\cdot 2\cdot 3\cdots n}\geq n.$$

This implies that $\lim_{n\to\infty}a_n=\infty$. Hence by the Divergence Test the series diverges.

**Cauchy Root Test**

*Theorem* (Cauchy Root Test). Suppose that $\sum_{n=1}^\infty a_n$ be a series with positive terms.

- If $\lim_{n\to\infty}\root n\of{a_n}=r<1$ then $\sum_{n=1}^\infty a_n$ converges.
- If $\lim_{n\to\infty}\root n\of{a_n}=r> 1$ then $\sum_{n=1}^\infty a_n$ diverges.
- If $\lim_{n\to\infty}\root n\of{a_n}=r=1$ then the test fails, i.e., the root test is inclusive.

*Example*. Test the convergence of the series $\sum_{n=1}^\infty\left(\frac{2n+3}{3n+2}\right)^n$.

*Solution*. \begin{align*}\lim_{n\to\infty}\root n\of{a_n}&=\lim_{n\to\infty}\root n\of{\left(\frac{2n+3}{3n+2}\right)^n}\\&=\lim_{n\to\infty}\frac{2n+3}{3n+2}\\&=\frac{2}{3}<1\end{align*}Hence by the Root Test the series converges.

**Comparison Test**

*Theorem* (Comparison Test). Suppose that $\sum_{n=1}^\infty a_n$ and $\sum_{n=1}^\infty b_n$ be series with positive terms.

- If $\sum_{n=1}^\infty b_n$ converges and $a_n\leq b_n$ for all $n$, then $\sum_{n=1}^\infty a_n$ also converges.
- If $\sum_{n=1}^\infty b_n$ diverges and $b_n\leq a_n$ for all $n$, then $\sum_{n=1}^\infty a_n$ also diverges.

*Remark*. For a convergent series we have the geometric series, whereas the harmonic series will serve as a divergent series. As other series are identified as either convergent or divergent, they may be used for the known series in this comparison test.

Example. Determine whether the series $\sum_{n=1}^\infty\frac{5}{2n^2+4n+3}$ converges.

*Solution*. Notice that $\frac{5}{2n^2+4n+3}<\frac{5}{n^2}$ for all $n$. Since $\sum_{n=1}\frac{1}{n^2}$ converges (it is a $p$-series with $p=2$), by the Comparison Test the series converges.

*Example*. Test the series $\sum_{n=1}^\infty\frac{\ln n}{n}$ for convergence or divergence.

*Solution*. $\left(\frac{\ln x}{x}\right)’=\frac{1-\ln x}{x}<0$ on $(e,\infty)$ i.e. $\frac{\ln n}{n}>\frac{1}{n}$ for all $n\geq 3$. Since $\sum_{n=1}^\infty\frac{1}{n}$ diverges (the harmonic series, also $p$-series with $p=1$), by the Comparison Test, the series diverges.

Example (The $p$ series). Let $p\leq 1$ Then

$\frac{1}{n}<\frac{1}{n^p}$ for all $n$, so by the Comparison Test

$\sum_{n=1}^\infty\frac{1}{n^p}$ is divergent for all $p\leq 1$.

**The Limit Comparison Test**

The Limit Comparison Test is a variation of the Comparison Test.

*Theorem* (The Limit Comparison Test). Suppose that $\sum_{n=1}^\infty a_n$ and $\sum_{n=1}^\infty b_b$ are series with positive terms. If $$\lim_{n\to\infty}\frac{a_n}{b_n}=c$$ where $c$ is a number and $c>0$, then either both series converge or both diverge.

*Example*. Test the series $\sum_{n=1}^\infty\frac{1}{2^n-1}$ for convergence or divergence.

*Solution*. Let $a_n=\frac{1}{2^n-1}$ and $b_n=\frac{1}{2^n}$. Then \begin{align*}\lim_{n\to\infty}\frac{a_n}{b_n}&=\lim_{n\to\infty}\frac{2^n}{2^n-1}\\&=\lim_{n\to\infty}\frac{1}{1-\frac{1}{2^n}}\\&=1>0\end{align*} Since $\sum_{n=1}^\infty\frac{1}{2^n}$ converges, so should $\sum_{n=1}^\infty\frac{1}{2^n-1}$ by the Limit Comparison Test.

*Example*. Test the series $\sum_{n=1}^\infty\frac{1}{\sqrt{n^2+1}}$ for convergence or divergence.

*Solution*. Let $a_n=\frac{1}{\sqrt{n^2+1}}$ and $b_n=\frac{1}{n}$. Then \begin{align*}\lim_{n\to\infty}\frac{a_n}{b_n}&=\lim_{n\to\infty}\frac{n}{\sqrt{n^2+1}}\\&=\lim_{n\to\infty}\frac{1}{\sqrt{1+\frac{1}{n^2}}}\\&=1>0\end{align*} Since $\sum_{n=1}^\infty\frac{1}{n}$ diverges, so should $\sum_{n=1}^\infty\frac{1}{\sqrt{n^2+1}}$ by the Limit Comparison Test.