Respuesta :

Answer:

False

Step-by-step explanation:

We have the serie:

[tex]\frac{1}{49}+ \frac{1}{64} + \frac{1}{81}+...[/tex]

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

Note that:

[tex]7^2 = 49\\\\8^2 = 64\\\\9^2 = 81[/tex]

Then we can write the series as:

[tex]\frac{1}{7^2}+ \frac{1}{8^2} + \frac{1}{9^2}+...[/tex]

Then:

[tex]\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}[/tex]

The series that have the form:

[tex]\sum_{n=1}^{\infty}\frac{1}{n^p}[/tex]

are known as "p-series". This type of series converges whenever [tex]p > 1[/tex].

In this case, [tex]p = 2[/tex] and [tex]2 > 1[/tex]. Then the series converges