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.
Best Answer
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 $\sum_{n=1}^\infty \frac{1}{n^p}$.