Convergence test from Demidovich: $\sum_{n=1}^{\infty} \frac{n^{n + \frac{1}{n}}}{(n + \frac{1}{n})^n}$

Question

I'm just learning how to test series for convergence and have encountered this series from the Demidovich's book and I can't really decide what criteria should I use. Could you please give me some hint(s)?

$$\sum_{n=1}^{\infty} \frac{n^{n + \frac{1}{n}}}{(n + \frac{1}{n})^n}$$

Answer 1

Hint: Since $n^{1/n} \to 1,$ you can forget about it. We're left with an $n$th term equal to

$$\frac{n^n}{(n+1/n)^n} = \left ( \frac{1}{1 + 1/n^2} \right )^n.$$

Answer 2

The $n^{th}$ root test looks tempting because of all the $n^{th}$ powers, but it only works when the rate of increase or decrease of the terms of the series is exponential, which is not the case here. Let's work out a couple of limits instead: $$\lim_{n\rightarrow\infty}n^{\frac1n}=\lim_{n\rightarrow\infty}\exp\left(\frac{\ln n}n\right)=\exp\left(\frac 00\right)=\lim_{n\rightarrow\infty}\exp\left(\frac{\frac1n}1\right)=\exp\left(\frac01\right)=e^0=1$$ And $$\begin{align}\lim_{n\rightarrow\infty}\left(1+\frac1{n^2}\right)^n&=\lim_{n\rightarrow\infty}\exp\left(n\ln\left(1+\frac1{n^2}\right)\right)\\ &=\lim_{n\rightarrow\infty}\exp\left(\frac{\ln\left(1+\frac1{n^2}\right)}{\frac1n}\right)=\exp\left(\frac00\right)\\ &=\lim_{n\rightarrow\infty}\exp\left(\frac{\frac{-\frac2{n^3}}{1+\frac1{n^2}}}{-\frac1{n^2}}\right)=\lim_{n\rightarrow\infty}\exp\left(\frac{\frac2n}{1+\frac1{n^2}}\right)\\ &=\exp\left(\frac0{1+0}\right)=e^0=1\end{align}$$ Then $$\lim_{n\rightarrow\infty}\frac{n^{n+\frac1n}}{\left(n+\frac1n\right)^n}=\lim_{n\rightarrow\infty}\frac{n^{\frac1n}}{\left(1+\frac1{n^2}\right)^n}=\frac11=1\ne0$$ So the sum diverges by the divergence test.