Question

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

Answer 1

False

Explanation:

We have the serie:

`(1)/(49)+ (1)/(64) + (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:

`(1)/(7^2)+ (1)/(8^2) + (1)/(9^2)+...`

Then:

`(1)/(7^2)+ (1)/(8^2) + (1)/(9^2)+... = \sum_(n=7)^(\infty)(1)/(n^2)\\\\\sum_(n=7)^(\infty)(1)/(n^2) = \sum_(n=1)^(\infty)(1)/((n+6)^2)`

The series that have the form:

`\sum_(n=1)^(\infty)(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