Does the following series converge or diverge: $\sum_{n=1}^{\infty}\frac{n!}{n^{n}}$?

Question

Which of the following series converge, and which diverge?

• $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{2}+1}$

• $\displaystyle\sum_{n=1}^{\infty}\frac{n!}{n^{n}}$

• $\displaystyle\sum_{n=1}^{\infty}\frac{1}{\sqrt{n^{2}+n}}$

MY ATTEMPT

The first one converges due to the comparison test. Indeed, one has \begin{align*} \sum_{n=1}^{\infty}\frac{1}{n^{2}+1} \leq \sum_{n=1}^{\infty}\frac{1}{n^{2}} \end{align*} where the last series is the $p$-series with $p = 2 > 1$.

The second series does converge due to the ratio test. Indeed, one has \begin{align*} \lim_{n\to\infty}\frac{(n+1)!}{(n+1)^{n+1}}\times\frac{n^{n}}{n!} = \lim_{n\to\infty}\left(\frac{n}{n+1}\right)^{n} = \lim_{n\to\infty}\left(1 - \frac{1}{n+1}\right)^{n} = e^{-1} < 1 \end{align*}

Finally, for the third series it suffices to notice that $n^{2} + n \leq 2n^{2}$. Thence we conclude that it diverges. Indeed, one has \begin{align*} \sum_{n=1}^{\infty}\frac{1}{\sqrt{n^{2}+n}} \geq \frac{1}{\sqrt{2}}\sum_{n=1}^{\infty}\frac{1}{n} \longrightarrow +\infty \end{align*}

Is the wording of my solutions good? Any comments and contributions are appreciated.

Answer 1

Most of this is correct and well-worded. Where you have used the phrase "harmonic series with $p = 2$", you really should say "$p$-series with $p = 2$. The harmonic series is $\sum_{n=1}^\infty \frac{1}{n}$, whereas $p$-series are series of the form $\sum_{n=1}^\infty \frac{1}{n^p}$.