Is this going to converge? $\sum_{n=1}^{\infty}\frac{1}{n^{n}}$ I have this series $\sum_{n=1}^{\infty}\frac{1}{n^{n}}$
is it converging, or diverging, what test do you use?
 A: $\frac{1}{n^n} \le \frac{1}{n^2}$ for $n\ge2$ so you can compare it with $\sum\frac{1}{n^2}$
A: Note that $n^n\ge n!>0$.  
Since $n^n>0$, the sequence of partial sums, $S_N=\sum_{n=1}^N \frac{1}{n^n}$ monotonically increases.
Moreover, we see that 

$$\left|\sum_{n=1}^N \frac{1}{n^n}\right|\le \sum_{n=1}^N\frac{1}{n!}<e$$

Inasmuch as the sequence of partial sums is monotonically increasing and bounded above (by $e$ in fact), it converges.  

Aside, the series is part of the Sophomore's Dream since it is equal to the integral 
$$\int_0^1 \frac{1}{x^x}\,dx=\sum_{n=1}^\infty \frac{1}{n^n}\approx 1.29129$$

We can obtain a crude upper bound for the series by noting that (see the Integral Test)
$$\begin{align}
\sum_{n=1}^\infty \frac{1}{n^n}&\le 1+\int_1^\infty \frac{1}{x^x}\,dx\\\\
&=1+\int_0^1 e^{-x\log(x)}\,dx\\\\
&\le 2-\frac1e \\\\
&\approx 1.632120559
\end{align}$$
This upper bound has an error of roughly than $26$% with the actual value of the series.
A: With an nth term with a power in the denominator, we can compare this series to a p-series. We must, however, pick the correct p-series.
The p-series test states that:
If the nth term of the sequence of your series is of the form $\frac {1}{n^p}$, then,
for all $p\gt 1$, the series converges, and
for all $p\le1$, the series diverges. 
Now we will use the direct comparison test, which states that:
If $0 \lt a_n \lt b_n$ for all n, then
If $\sum_{n=1}^{\infty}b_n$ converges, $\sum_{n=1}^{\infty}a_n$ converges.
If $\sum_{n=1}^{\infty}a_n$ diverges, $\sum_{n=1}^{\infty}b_n$ diverges.
We know that $\frac{1}{n^n} \le \frac{1}{n^2}$ for all $n\ge1$. You can easily test this by plugging in values for $n$ into both formulas. Thus, let $b_n = \frac {1}{n^2}$ and $a_n = \frac {1}{n^n}$.
By the p-series test, we know that the series $\sum_{n=1}^{\infty}\frac{1}{n^2}$ converges: $p = 2 \gt 1$.
$\sum_{n=1}^{\infty}b_n$ converges.
If $\sum_{n=1}^{\infty}b_n$ converges, then $\sum_{n=1}^{\infty}a_n$ converges.
Thus, $\sum_{n=1}^{\infty}\frac{1}{n^n}$ converges.
Hoorah! Finally, we conclude that because the series $\sum_{n=1}^{\infty}\frac{1}{n^2}$ converges by the p-series test, the series $\sum_{n=1}^{\infty}\frac{1}{n^n}$ converges by the direct comparison test.

Note: I chose to compare $\frac{1}{n^n}$ to $\frac{1}{n^2}$ instead of $\frac{1}{n}$ for a good reason. First notice that $\frac {1}{n^n} \le \frac 1n$ for all $n \ge 1$, so $b_n = \frac 1n$ and $a_n =\frac {1}{n^n}$.  $\sum_{n=1}^{\infty}\frac{1}{n}$ is the harmonic series, and by the p-series test we know that this diverges. Here is the crucial part: $b_n$ diverges. We cannot deduce anything from the direct comparison test if $b_n$ diverges. Thus, I went with $\frac {1}{n^2}$ as $b_n$, instead. This converges by the p-series test, and now you see that we can deduce information from the direct comparison test if $b_n$ converges - namely, that $a_n$ will converge as well.
A: The inequality for $n\ge1$
$$
n^n\ge\left(n-\tfrac12\right)\left(n+\tfrac12\right)
$$
can be verified for $n=1$ and for $n\ge2$
$$
n^n\ge n^2\ge n^2-\tfrac14=\left(n-\tfrac12\right)\left(n+\tfrac12\right)
$$
Therefore,
$$
\begin{align}
\sum_{n=1}^\infty\frac1{n^n}
&\le\sum_{n=1}^\infty\frac1{\left(n-\frac12\right)\left(n+\frac12\right)}\\
&=\sum_{n=1}^\infty\left[\frac1{n-\frac12}-\frac1{n+\frac12}\right]\\[6pt]
&=2
\end{align}
$$
A: Let $a_n=\frac{1}{n^n}$. 
Then 
$$|a_n|^{1/n}=\frac{1}{n} \to 0<1.$$
The root-test shows that the series converges.
