How do you find the interval of convergence $Sigma x^n/n^2$ from $n=[1,oo)$ ?

Question

25388 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-05T05:56:05

The interval of convergence is $-1<=x<=1$

Use the Ratio test to compute the convergence interval

$a_n=x^n/n^2$

$|a_(n+1)/a_n|=|x^(n+1)/(n+1)^2*n^2/x^n|$

$=(n^2x)/(n+1)^2$

Calculate the limit as $n->oo$

Therefore,

$lim_(n->oo)|(n^2x)/(n+1)^2|$

$=|x|*lim_(n->oo)|(n^2)/(n+1)^2|$

$=|x|*1$

$=|x|$

The series converges for $|x|<1$

Now, check the convergence when, $|x|=1$

When $x=1$

$sum_1^oo1^n/n^2$

By the $" p-series test"$ , $p=2$ and $p>1$ , the series converges

When $x=-1$

$sum_1^oo(-1)^n/n^2=sum_1^oo|(-1)^n/n^2|=sum_1^oo1/n^2$

By the $" p-series test"$ , $p=2$ and $p>1$ , the series converges

The series converges absolutely

How do you find the interval of convergence Sigma x^n/n^2Sigma x^n/n^2 from n=[1,oo)n=[1,oo)?