Convergence of the series $\frac1{n^n}$ Does the series $\frac1{n^n}$ converges or diverges .
By comparison test I can claim that $\frac1{n^n} < \frac1n$ , and since series $\frac1n$ appears divergent , so the series $\frac1{n^n}$ also diverges.
However by Cauchy root test the limit of $\frac1{n^n}$ appears to be $0<1$ which suggests the convergence of the series 
 A: We cannot conclude that since $\frac{1}{n^n} < \frac{1}{n}$ then $\sum_{n=1}^\infty \frac{1}{n^n}$ diverges.
If $\frac{1}{n^n} > \frac{1}{n}$ (which is not true), then we can conclude that $\sum_{n=1}^\infty \frac{1}{n^n}$ diverges.
You have used Cauchy root test to conclude that it converges.
If we want to use comparison test, notice that $$\frac{1}{n^n} \leq\frac{1}{n^2}$$
and since $\sum_{n=1}^\infty \frac1{n^2}$ conveges, hence $\sum_{n=1}^\infty \frac1{n^n}$ converges.
A: The comparison test tells you that if $$a_n> b_n$$ and $$\sum_{n}a_n$$ converges, then $$\sum_n b_n$$ also converges.
It does not say that if $\sum_{n} a_n$ diverges then $\sum_{n} b_n$ diverges. If it did, then because $\frac{1}{2^n} < 1$, you could conclude that $\sum \frac1{2^n}$ also diverges.
In fact, you would need the inequality reversed, so if $a_n<b_n$ and $\sum a_n$ diverges, then $\sum b_n$ also diverges.
So, your logic is incorrect.

In fact, the series $\sum\frac{1}{n^n}$ converges by the comparison test with $\frac{1}{2^n}$ or with $\frac{1}{n^2}$.
A: A series converges if it has a converging upper bound and it diverges if it has a diverging lower bound. In other cases, you cannot conclude.
In this particular case, you can for instance exploit
$$\frac1{n^n}\le \frac1{2^n}$$ for $n\ge2$. Actually, the sequence converges extremely quickly, super-exponentially.
