Check if the series $\sum^{\infty}_{n=1} \frac{n^{n+\frac{1}{n}}}{(n+\frac{1}{n})^n}$ converges I am totally confused about the series:
$$\sum^{\infty}_{n=1} \frac{n^{n+\frac{1}{n}}}{(n+\frac{1}{n})^n}$$
it seems like root test should be ok, but I don't know how to apply it here (if it is possible at all)
 A: You can simplify by $n^n$ and get
$$\frac{n^{1/n}}{\left(1+\frac1{n^2}\right)^n}.$$
Both the numerator and denominator are known to tend to $1$, hence the sum diverges.
A: Hint:
Write,
$$ \frac{n^{n+\frac{1}{n}}}{(n+\frac{1}{n})^n}$$
$$=\frac{n^{n}n^{\frac{1}{n}}}{(n+\frac{1}{n})^n}$$
$$=(\frac{n}{n+\frac{1}{n}})^n n^{\frac{1}{n}}$$
$$=(\frac{1}{1+\frac{1}{n^2}})^n n^{\frac{1}{n}}$$ 
$$=e^{n \ln \left(\frac{1}{1+\frac{1}{n^2}}\right)} e^{\frac{1}{n} \ln n}$$
$$=e^{\frac{\ln \left(\frac{1}{1+\frac{1}{n^2}}\right)}{\frac{1}{n}}} e^{\frac{1}{n} \ln n}$$
Evaluate this as $n \to \infty$. (Perhaps using l'Hopitals, and product of limits) 
A: Root test becomes inconclusive (=1) eventually. Consider just $$\lim_{n\to\infty}a_n=\lim_{n\to\infty}\dfrac{n^{n+\frac{1}{n}}}{(n+\frac{1}{n})^n}=\lim_{n\to\infty}\dfrac{n^{n}\cdot n^{\frac{1}{n}}}{\underbrace{n^n}_{n^{\text{th}}\text{ power of } n}+\underbrace{\binom{n}{1}n^{n-1}\frac{1}{n}}_{(n-1)^{\text{th}}\text{ power of } n}+\underbrace{\binom{n}{2}n^{n-2}\frac{1}{n^2}}_{(n-2)^{\text{th}}\text{ power of } n}+\ldots \binom{n}{n-1}n^{1}\frac{1}{n^{n-1}}}$$ $$=\lim_{n\to\infty}\dfrac{1}{1+\binom{n}{1}n^{-1}\frac{1}{n}+\binom{n}{2}n^{-2}\frac{1}{n^2}+\ldots \binom{n}{n-1}n^{1-n}\frac{1}{n^{n-1}}}=\dfrac{1}{1+0+0+\ldots+0}=1$$
where we used the binomial expansion, we used the fact that $\lim\limits_{x\to\infty}x^{\frac{1}{x}}=1$, and in the the third step, I multiplied by $\dfrac{\frac{1}{n^n}}{\frac{1}{n^n}}=1$.
A: This series diverges trivially since
\begin{align}
\ln\frac{n^{n+\tfrac1n}}{\Bigl(n+\cfrac1n\Bigr)^n}&=\Bigl(n+\frac1n\Bigr)\ln n- n\ln\Bigl(n+\frac1n\Bigr)=\Bigl(n+\frac1n\Bigr)\ln n- n\ln n-n\ln\Bigl(1+\frac1{n^2}\Bigr)\\
&=\frac{\ln n}n-n\ln\Bigl(1+\frac1{n^2}\Bigr)=\frac{\ln n}n-n\biggl(\frac1{n^2}+o\Bigl(\frac1{n^2}\Bigr)\biggr)=\frac{\ln n}n-\frac1n-o\Bigl(\frac1n\Bigr)\to 0
\end{align}
so the general term of the series tends to $1$.
