# Interval of Convergence Q: Find the interval of convergence of this power series: 2x + `x^(2)` + `(8)/(9)` `x^(3)` + `x^(4)` +... = `sum_(n=1)^oo` `<SPAN class=AM></SPAN> `...

Interval of Convergence

Q: Find the interval of convergence of this power series:

2x + `x^(2)` + `(8)/(9)` `x^(3)` + `x^(4)` +... = `sum_(n=1)^oo` `<SPAN class=AM></SPAN> ` `(2^n)/(n^2)` `x^(n)`

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Q: Find the interval of convergence of this power series:

2x + `x^(2)` + `(8)/(9)` `x^(3)` + `x^(4)` + ... = `sum_(n=1)^oo` `(2^n)/(n^2)` `x^(n)`

This is power series around 0 so it will converge for `x=0`. Now we need to find the radius of convergence `r` which tells us how far away from 0 will the series converge. There are several different ways to find convergence radius and I will show you one of them.

`r=lim_(n->oo)1/root(n)(|a_n|)`

In your case `a_n=2^n/n^2` so we have

`r=lim_(n->oo)1/root(n)(2^n/n^2)=lim_(n->oo)1/(root(n)(2^n)/root(n)(n^2))=1/(2/1)=1/2`

` `` ` So radius of convergence is `r=1/2` which means that your series will converge for `|x-0|<1/2=>|x|<1/2`.

**The series is convergent for** `|x|<1/2` (`-1/2<x<1/2` )