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