End points interval of convergence

Question

I am trying to evaluate the end points of an interval of convergence of the series: $$\sum_{n=1}^{\infty}\frac{((n+1)x)^n}{n^{n+1}}$$ Applying root test: $$\lim_{n \to \infty}\sqrt[n]{\left|\frac{(n+1)x)^n}{n^{n+1}}\right|}$$ $$=\lim_{n \to \infty}\left|\frac{(n+1)x}{n^{1+\frac{1}{n}}}\right|=\left|x\right|$$ But when I plug in $x=1$ we will get the same as above and root test will return $1$ again, which isn't much help so I was wandering if I went wrong somewhere or if there is a better way to solve this problem? Thanks

Answer 1

From $$\lim\limits_{n\to\infty}{\left(1+\dfrac{1}{n}\right)^n}=e$$ we have that $$\dfrac{(n+1)^n}{n^{n+1}}=\dfrac{1}{n}\cdot\dfrac{(n+1)^n}{n^{n}}=\dfrac{1}{n}\cdot\left(1+\dfrac{1}{n}\right)^n=O\left(\dfrac{1}{n}\right),\;\;n\to\infty.$$

Answer 2

As $n\to\infty$, $\displaystyle \frac{(n+1)^n}{n^{n+1}} \approx \frac e n $ so your series should have end point of convergence like of $ \displaystyle \sum \frac{x^n}{n}$.
In your attempt of solution, if $$\lim_{n \to \infty}\left|\frac{(n+1)x}{n^{1+\frac{1}{n}}}\right| < 1$$ then the series converges absolutely and $$\lim_{n\to\infty}\frac{n+1}{n^{1 + \frac 1 n}} = 1$$ that will give you $|x| < 1$