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

Accepted Solution

Answer:FalseStep-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 seriesNote 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