Chapter 10, Problem 50RE

To determine

To find: The series converges absolutely, conditionally or diverges.

Answer to Problem 50RE

  ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{\sqrt{n\left(n+1\right)\left(n+2\right)}}}$ convergesabsolutely.

Explanation of Solution

Given information:

The given series is,

  ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{\sqrt{n\left(n+1\right)\left(n+2\right)}}}$

Formula used:

p-test series Test, ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^p}}$

Calculation:

Using direct comparison test: Now, consider the series ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{\sqrt{n^3}}}$ by p-test series Test, ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^p}}$ converges if $p>1$ , and diverges if $p\le 1$

It is of the form ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^p}}$ where $p=\frac{3}{2}$ . Therefore, this series converges, since p > 1.

Hence, the limit is positive and ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{\sqrt{n^3}}}$ converges

Since,

  $\sqrt{n\left(n+1\right)\left(n+2\right)}>\sqrt{n^3}\frac{1}{\sqrt{n\left(n+1\right)\left(n+2\right)}}<\frac{1}{\sqrt{n^3}}$

And

  ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{\sqrt{n^3}}}$ converges.

Therefore,

  ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{\sqrt{n\left(n+1\right)\left(n+2\right)}}}$ converges absolutely.

Conclusion:

  ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{\sqrt{n\left(n+1\right)\left(n+2\right)}}}$ converges absolutely.