Question

Use the special comparison test to find whether the following series converge or diverge. $$\sum _{n=0}^{\mathrm{\infty }}\frac{n(n+1)}{(n+2{)}^2(n+3)}$$

Short answer

The series converges.

Step-by-step solution

- Identify the type of series

Recognize that the given series $\frac{n(n+1)}{(n+2{)}^2(n+3)}$ can be studied using the comparison test because it has rational terms.

- Compare with a simpler series

To apply the comparison test, compare $a_n=\frac{n(n+1)}{(n+2{)}^2(n+3)}$ with a simpler series. Consider $b_n=\frac{1}{n^2}$ for large $n$.

- Simplify the original term

Simplify $a_n$: As $n\to \mathrm{\infty }$, $a_n\approx \frac{n^2}{n^2\times n}=\frac{1}{n^2}$.

- Establish inequality for comparison

For sufficiently large $n$, $a_n$ behaves similarly to $b_n=\frac{1}{n^2}$. Therefore, $0<a_n<C\times b_n$ for some constant $C>0$.

- Determine if the comparison series converges

Observe that $\frac{1}{n^2}$ is a $p$-series with $p=2$, which is known to converge because $p>1$.

- Apply the limit comparison test

To solidify the comparison, use the limit comparison test. Calculate ${\text{lim}}_{n\to \mathrm{\infty }}\frac{a_n}{b_n}={\text{lim}}_{n\to \mathrm{\infty }}\frac{\frac{n(n+1)}{(n+2{)}^2(n+3)}}{\frac{1}{n^2}}={\text{lim}}_{n\to \mathrm{\infty }}\frac{n^3+n^2}{n^4}={\text{lim}}_{n\to \mathrm{\infty }}\frac{n+1}{n^2}=1$. Since the limit is a positive finite value, the behavior of $\frac{n(n+1)}{(n+2{)}^2(n+3)}$ is similar to $\frac{1}{n^2}$.

- Conclude the series behavior

Since $\frac{1}{n^2}$ converges and $\frac{n(n+1)}{(n+2{)}^2(n+3)}$ behaves similarly to $\frac{1}{n^2}$ via the comparison test, the original series $\frac{n(n+1)}{(n+2{)}^2(n+3)}$ also converges.

Key concepts

Series Convergence

When we talk about series convergence, we're referring to whether the sum of the infinite series approaches a finite value as the number of terms increases indefinitely. To decide if a series converges or diverges, we use various tests and methods.

For our series, $\sum _{n=0}^{\mathrm{\infty }}\frac{n(n+1)}{(n+2{)}^2(n+3)}$, we need to determine if the sum of its terms increases without bound (diverges) or stabilizes to a finite value (converges).

Comparison Test

The comparison test is a useful tool to check whether a series converges. It involves comparing the given series to another series whose convergence behavior is known.

For our series, $\frac{n(n+1)}{(n+2{)}^2(n+3)}$, we compare it to a simpler series, $\frac{1}{n^2}$, called the 'comparison series'. If $\frac{1}{n^2}$ is larger and converges, our series likely converges too.

• If $\sum b_n$ converges and $\sum a_n\le C{b}_n$ for some constant \C > 0\, then $\sum a_n$ also converges
• If $\sum b_n$ diverges and $\sum a_n\ge C{b}_n$ for some constant \C > 0\, then $\sum a_n$ also diverges

Limit Comparison Test

The limit comparison test is an extension of the comparison test. It helps when the straightforward comparison isn't clear. We take the limit of the ratio of the terms of our series to the comparator series. ${\text{lim}}_{n\to \mathrm{\infty }}\frac{a_n}{b_n}$. If this limit is a positive finite number, both series converge or both diverge together.
In our example, this limit calculation would be
${\text{lim}}_{n\to \mathrm{\infty }}\frac{\frac{n(n+1)}{(n+2{)}^2(n+3)}}{\frac{1}{n^2}}$ = \frac{n^3 + n^2}{n^4} \rightarrow 1
This positive finite limit confirms that if $\frac{1}{n^2}$ converges, so does $\frac{n(n+1)}{(n+2{)}^2(n+3)}$

P-Series

A p-series is a series of the form $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^p}$, where \p\ is a positive real number. These series have well-known convergence properties.

Specifically:

• For \p \leq 1\, the series diverges
• For \p > 1\, the series converges
  In our problem, $\frac{1}{n^2}$ is a p-series with \p = 2\, which converges. This helps us determine the behavior of $\frac{n(n+1)}{(n+2{)}^2(n+3)}$.

Infinite Series

An infinite series is a sum of infinitely many terms. Understanding whether an infinite series converges or diverges is essential in math and applied sciences.

In the case of $\sum _{n=0}^{\mathrm{\infty }}\frac{n(n+1)}{(n+2{)}^2(n+3)}$, we used tests to compare its behavior to simpler, well-understood series. This helps us understand the sum's behavior as the number of terms becomes very large.
Using the comparison, limit comparison, and knowledge of p-series convergence, we concluded that our series converges due to its association with the convergent p-series $\frac{1}{n^2}$.