Direct comparison test: Suppose $\displaystyle\sum_{n=1}^\infty a_n$ and $\displaystyle\sum_{n=1}^\infty b_n$ are series with positive terms.

• If $\displaystyle\sum_{n=1}^\infty b_n$ converges and $a_n \le b_n$, then $\displaystyle\sum_{n=1}^\infty a_n$ is also convergent.
• If $\displaystyle\sum_{n=1}^\infty a_n$ diverges and $a_n \le b_n$, then $\displaystyle\sum_{n=1}^\infty b_n$ is also divergent.

Limit comparison test: Suppose $\displaystyle\sum_{n=1}^\infty a_n$ and $\displaystyle\sum_{n=1}^\infty b_n$ are series with positive terms.

• If $\displaystyle\lim_{n\to\infty}\frac{a_n}{b_n} = c$, where $c$ is a finite number such that $c > 0$, then both series converge or both series diverge.
• It’s typically a good idea to use the limit comparison test with series that behave similarly for large $n$. (The next example demonstrates this.)

an example

<aside>

Does $\displaystyle\sum_{n=1}^\infty\frac{1}{n^2-1}$ converge or diverge?

</aside>

We can use the limit comparison test to $\displaystyle\sum_{n=1}^\infty\frac{1}{n^2}$, since for large $n$, $\displaystyle\frac{1}{n^2-1} \approx \frac{1}{n^2}$.

So let’s compute $\displaystyle\lim_{n\to\infty}\frac{a_n}{b_n}$:

$$ \displaystyle\lim_{n\to\infty}\frac{a_n}{b_n} = \lim_{n\to\infty}\frac{1/(n^2-1)}{1/n^2}=\lim_{n\to\infty}\frac{n^2}{n^2-1} =\lim_{n\to\infty}\frac{1}{1-1/n^2} = 1 $$

Since $\displaystyle\sum_{n=1}^\infty\frac{1}{n^2}$ converges (since it is a $p$-series with $p > 1$), the series $\displaystyle\sum_{n=1}^\infty\frac{1}{n^2-1}$ also converges.

another example

<aside>

Does $\displaystyle\sum_{n=1}^\infty\frac{2+(-1)^n}{n\sqrt{n}}$ converge or diverge?

</aside>

We can use the direct comparison test here with these two choices for the series:

$$

\displaystyle\sum_{n=1}^\infty a_n =\displaystyle\sum_{n=1}^\infty\frac{2+(-1)^n}{n\sqrt{n}} $$

$$

\displaystyle\sum_{n=1}^\infty b_n =\displaystyle\sum_{n=1}^\infty\frac{3}{n^{3/2}} $$

Then $\displaystyle\sum_{n=1}^\infty a_n \le \displaystyle\sum_{n=1}^\infty b_n < \infty$, so the series converges.

yet another example

<aside>

Does $\displaystyle\sum_{n=1}^\infty\frac{n+2}{(n+1)^3}$ converge or diverge?

</aside>