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.
 A: Observe that $$\lim_{n \to \infty }\sqrt[n]{\frac{3^{n}+(-2)^{n}}{n}} = 3$$ so by Cauchy -Hadamard the radius of convergence of your series is $\frac{1}{3}$, i.e. the series is convergent for $|x+1|<\frac{1}{3} $ and doesn't converge for $|x+1|>\frac{1}{3} $
A: First way. The radius of convergence can be found using the ratio test, since
$$
\frac{a_{n+1}}{a_{n}}=\frac{n}{n+1}\cdot \frac{3^{n+1}+(-2)^{n+1}}{3^{n}+(-2)^{n}}=3\cdot
\frac{1}{1+\frac{1}{n}}\cdot \frac{1+\left(-\frac{2}{3}\right)^{n}}{1+\left(-\frac{2}{3}\right)^{n+1}}\,\to\, 3.
$$
Second way. Here you can use the root test as well:
If $\,\,\limsup_{n\to\infty} |a_n|^{1/n}=\ell$, then the radius of convergence is $r=1/\ell$.
Here
$$
\limsup_{n\to\infty}\left|\frac{(3^n+(-2)^n )}{n}\right|^{1/n}=3,
$$
and hence $r=1/3$.
In fact, here the sequence $\left|\frac{(3^n+(-2)^n )}{n}\right|^{1/n}$ converges to $3$ since
$$
\frac{1}{3n}\cdot 3^n=\frac{1}{n} 3^{n-1}=\frac{3^n+(-2)^n }{n}\le \frac{3}{n}\cdot 3^n,
$$
and 
$\big(\frac{1}{3n}\cdot 3^n\big)^{1/n}\to 3$ and $\big(\frac{3}{n}\cdot 3^n\big)^{1/n}\to 3$ as well.
A: $lim_{n->\infty} \frac{n(3^{n+1}+(-2)^{n+1} )}{(n+1)(3^n+(-2)^n)}$
$=lim_{n->\infty} \frac{3^{n+1}n(1+(-2/3)^{n+1} )}{3^nn(1+1/n)(1+(-2/3)^n)}$
$=lim_{n->\infty} \frac{3(1+(-2/3)^{n+1} )}{(1+1/n)(1+(-2/3)^n)}=\frac{3(1+0)}{1+0)(1+0)}=3$
