Does the series $\sum_{n=1}^\infty\frac{n^{\sqrt{n}}}{n!}$ converge? Immediately I recognize that there's a factorial and I use the ratio test to try and solve it:
$$\lim_{n \rightarrow \infty}\left|\frac{{(n+1)}^{\sqrt{n+1}}}{(n+1)!}\cdot\frac{n!}{n^{\sqrt{n}}}\right|=\lim_{n \rightarrow \infty}\left|\frac{{(n+1)}^{\sqrt{n+1}-1}}{n^{\sqrt{n}}}\right|$$
At this point I'm not sure how to evaluate the limit. The answer key says the limit is 0, but how is it getting that? Is there an easier way to see that the series converges?
 A: $\displaystyle\sum_{n=1}^{\infty}\frac{2^n}{n!}$ converges by the Ratio Test $\;\;$ (since $\displaystyle\lim_{n\to\infty}\frac{2^{n+1}}{(n+1)!}\cdot\frac{n!}{2^n}=\lim_{n\to\infty}\frac{2}{n+1}=0<1)$; 
and $\displaystyle\ln n\le \sqrt{n}\ln 2\implies \sqrt{n}\ln n\le n\ln 2\implies n^{\sqrt{n}}\le 2^n\implies\frac{n^{\sqrt{n}}}{n!}\le\frac{2^n}{n!}$ for $n$ large,
so $\displaystyle\sum_{n=1}^{\infty}\frac{n^{\sqrt{n}}}{n!}$ converges by the Comparison Test.
A: $n!\geq \left(\frac{n}{e}\right)^n$ is simple to prove, and it trivially gives that the series is convergent.
A: Note that we can write
$$\begin{align}
\frac{(n+1)^{\sqrt{n+1}}}{n^{\sqrt{n}}}&=e^{(\sqrt{n+1}-\sqrt{n})\log(n)+\sqrt{n+1}(\log(n+1)-\log(n))}\\\\
&=e^{\left(\frac{\log(n)}{\sqrt{n+1}+\sqrt{n}}\right)}e^{\sqrt{n+1}\log\left(1+\frac1n\right)}\\\\
&=e^{\left(\frac{\log(n)}{\sqrt{n+1}+\sqrt{n}}\right)}e^{\sqrt{n+1}\,O\left(\frac1n\right)}\\\\
&\to 1\,\,\text{as}\,\,n\to \infty
\end{align}$$
A: You can evuluate limit as follows:
$$\lim _{ n\rightarrow \infty  } \left| \frac { { (n+1) }^{ \sqrt { n+1 }  } }{ (n+1)! } \cdot \frac { n! }{ n^{ n } }  \right| =\lim _{ n\rightarrow \infty  } \left| \frac { { (n+1) }^{ \sqrt { n+1 } -1 } }{ n^{ n } }  \right| =\lim _{ n\rightarrow \infty  } \left| \frac { { \left( n+1 \right)  }^{ \frac { n }{ \sqrt { n+1 } +1 }  } }{ { n }^{ n } }  \right| =\\ =\lim _{ n\rightarrow \infty  } \left| \left[ { { \left( 1+\frac { 1 }{ n }  \right)  }^{ n } } \right] ^{ \frac { 1 }{ \sqrt { n+1 } +1 }  } \right| =\lim _{ n\rightarrow \infty  }{ e } ^{ \frac { 1 }{ \sqrt { n+1 } +1 }  }=1$$ 
but we know in case $1$ test doens't say about converges or diverges
A: Consider $$A_n=\frac{(n+1)^{\sqrt{n+1}}}{n^{\sqrt{n}}}$$ Take logarithms $$\log(A_n)={\sqrt{n+1}}\log(n+1)-{\sqrt{n}}\log(n)$$ $$\log(A_n)={\sqrt{n}} \sqrt{1+\frac 1 n}\left(\log(n)+\log(1+\frac 1 n)\right)-{\sqrt{n}}\log(n)$$ Now, let us use for large values of $n$ $$\sqrt{1+\frac 1 n}=1+\frac{1}{2 n}+O\left(\frac{1}{n^2}\right)$$ $$\log(1+\frac 1 n)=\frac{1}{n}+O\left(\frac{1}{n^2}\right)$$ and replace. This gives $$\log(A_n)=\frac{1+\frac{1}{2}\log (n)}{\sqrt{n}}+O\left(\frac{1}{n^{3/2}}\right)$$ So, when $n\to \infty$, $\log(A_n)\to 0$ and $A_n\to 1$.
