Determine for what value of $x$ the series converges $∑_{n=1}^∞ \frac{(3^n+(-2)^n )}{n} (x+1)^n $ Determine for what value of $x$ the series converges
$∑_{n=1}^∞ \frac{(3^n+(-2)^n )}{n} (x+1)^n $
Here is what I got
Using the ratiotest, I got
$D_n =\frac{\frac{(3^{n+1}+(-2)^{n+1} )}{n+1}(x+1)^{n+1}}{\frac {(3^n+(-2)^n )}{n} (x+1)^n}$
$=\frac{n(3^{n+1}+(-2)^{n+1} )}{(n+1)(3^n+(-2)^n)}(x+1) ->3(x+1) <1$ if $x<-2/3$
Is this correct?
I use maple to calculate $lim_{n->\infty} \frac{n(3^{n+1}+(-2)^{n+1} )}{(n+1)(3^n+(-2)^n)}=3$ but I don't know why they got this either.

$lim_{n->\infty} \frac{n(3^{n+1}+(-2)^{n+1} )}{(n+1)(3^n+(-2)^n)}$
$=lim_{n->\infty} \frac{3^{n+1}n(1+(-2/3)^{n+1} )}{3^nn(1+1/n)(1+(-2/3)^n)}$
$=lim_{n->\infty} \frac{3(1+(-2/3)^{n+1} )}{(1+1/n)(1+(-2/3)^n)}=\frac{3(1+0)}{1+0)(1+0)}=3$
