Convergence of series $\sum \limits^{\infty }_{n=1}\frac{n^{(n+\frac{1}{n} )}}{(n+\frac{1}{n} )^{\frac{1}{n} }}$

Question

i need help for find method or methods for solve this series and find the convergence. I very appreciate for any help and yours comments. $$\sum \limits^{\infty }_{n=1}\frac{n^{(n+\frac{1}{n} )}}{(n+\frac{1}{n} )^{\frac{1}{n} }} $$

Answer 1

That series is going to diverge because the numerator is close to $n^n$ while the denominator is close to $\sqrt[n]{n}$. Using the $n$-th term test for divergence:

$$ \lim_{n\rightarrow\infty}\frac{n^{(n+\frac{1}{n} )}}{(n+\frac{1}{n} )^{\frac{1}{n} }} $$

We can take the exponential and natural log to find that this limit equals $$ e^{\lim_{n\rightarrow\infty}(n+\frac{1}{n})\ln n-\frac{1}{n}\ln(n+\frac{1}{n})} $$

Consider the exponent: $$ \lim_{n\rightarrow\infty}(n+\frac{1}{n})\ln n-\frac{1}{n}\ln(n+\frac{1}{n}) $$

By rearranging, we have that this is $$ \lim_{n\rightarrow\infty}n\ln n+\frac{1}{n}(\ln n-\ln(n+\frac{1}{n})) $$

Since $\ln n-\ln(n+\frac{1}{n})=\ln\frac{n}{n+\frac{1}{n}}$, we see that $\frac{n}{n+\frac{1}{n}}$ approaches $1$ as $n$ grows, so this approaches $\ln(1)=0$. Moreover, the coefficient of $\frac{1}{n}$ also appraches zero, so this means that the second term in the limit above is

$$ \lim_{n\rightarrow\infty}n\ln n $$ which approaches infinity. Therefore, the limit of the $n$-th terms is $e^{\infty}$ (abusing notation), and so the series diverges.

Answer 2

For $n\ge 2$ we have $$\left(n+\frac{1}{n}\right)^{\frac{1}{n}}< n+\frac{1}{n}< n+1$$ Also, $$n^2>n+1$$ Then $$\frac{n^{n+\frac{1}{n}}}{\left(n+\frac{1}{n}\right)^{\frac{1}{n}}}>\frac{n^n}{n+1}>\frac{n^{n-2}(n+1)}{n+1}=n^{n-2}$$ It follows $$\sum_{n=1}^{\infty}\frac{n^{n+\frac{1}{n}}}{\left(n+\frac{1}{n}\right)^{\frac{1}{n}}}$$ diverges

Answer 3

The general term is divergent so the series diverges:

$\frac{n^{(n+\frac{1}{n} )}}{(n+\frac{1}{n} )^{\frac{1}{n} }}\geq \frac{n^{n}}{(2n)^{\frac{1}{n}}}=\left ( \frac{1}{2^{\frac{1}{n}}} \right )\left ( \frac{n^{n}}{n^{\frac{1}{n}}} \right )\rightarrow \infty $