find interval of convergence for
x^((n^2+2)/(n+3)), n=1-∞

Question

find interval of convergence for
x^((n^2+2)/(n+3)), n=1->∞

mr.math · Answer

Is it $\large x^{\frac{n^2+2}{n+3}}$?

mr.math · Answer

If so, then you can use the ratio test.
$$\large |{a_{n+1} \over a_n}|= \large |{x^{n^2+2n+3 \over n+4} \over x^{n^2+3 \over n+3}}|$$
$${n^2+2n+3 \over n+4}-{n^2+2 \over n+3}=({11 \over n+4}-{11 \over n+3}+1)\to 1 \text{ as } n \to \infty$$
Thus   $\large \lim_{n\to \infty}  |{a_{n+1} \over a_n}|=x.$

mr.math · Answer

That means that the series converges for $-1<x<1$. You need now to test the end points -1 and 1.

anonymous · Answer

i got the same result, but in my book says: absolute convergent for 0<x<1, divergent for x>=1

asnaseer · Answer

I would suggest that your book is wrong - @Mr.Math is definitely correct here. @Mr.Math made just one "typo" error where he has $x^{\frac{n^2+3}{n+3}}$ in the denominator of the ratio test instead of $x^{\frac{n^2+2}{n+3}}$ - i.e. should be $n^2+2$ not $n^2+3$, but that error was corrected in the next line.