Question

(10CQ) The series 1/49+1/64+1/81+ ... is divergent.

Answer 1

Answer:

False

Step-by-step explanation:

We have the serie:

$\frac{1}{49}+ \frac{1}{64} + \frac{1}{81}+...$

To test whether the series converges or diverges first we must find the rule of the series

Note that:

$7^2 = 49\\\\8^2 = 64\\\\9^2 = 81$

Then we can write the series as:

$\frac{1}{7^2}+ \frac{1}{8^2} + \frac{1}{9^2}+...$

Then:

$\frac{1}{7^2}+ \frac{1}{8^2} + \frac{1}{9^2}+... = \sum_{n=7}^{\infty}\frac{1}{n^2}\\\\\sum_{n=7}^{\infty}\frac{1}{n^2} = \sum_{n=1}^{\infty}\frac{1}{(n+6)^2}$

The series that have the form:

$\sum_{n=1}^{\infty}\frac{1}{n^p}$

are known as "p-series". This type of series converges whenever $p > 1$.

In this case, $p = 2$ and $2 > 1$. Then the series converges