Convergence of $\sum_{n=0}^\infty \frac{1}{n^n}$

Question

I'm stuck at deciding wether or not $\sum_{n=0}^\infty \frac{1}{n^n}$ converges.The sequence itself is a zero sequence and the root test seems to pass, but how can that be since for n=0 we would have have to deal with point of singularity.

The quantity, $0^0$ is generally set by convention to be $1$. — Baby Dragon, Apr 23 '13 at 21:12
Your question seems to ask about $0^0$, rather than the convergence of $\sum_{n=1}^\infty n^{-n}$, which you mentioned you already know how to prove using the root test. Your question is then a duplicate of http://math.stackexchange.com/questions/11150/zero-to-zero-power and I have closed it. — Eric Naslund, Apr 23 '13 at 21:22

score 1 · Accepted Answer · answered Apr 23 '13 at 21:12

The convergence of a series does not depend on any finite number of initial terms. In this case, $\frac{1}{0^0}$ has more than one interpretation (including undefined), but should not affect your answer. Most likely whoever wrote the problem intended to start with $n=1$.

score 1 · Answer 2 · answered Apr 23 '13 at 21:12

1

Single summands (as long as defined) do not interferer with convergence. And by the way, $\frac1{0^0}=\frac 1 1=1$.

answered Apr 23 '13 at 21:12

Hagen von Eitzen

374,180

score 0 · Answer 3 · answered Apr 23 '13 at 21:11

0

Hint: Use the comparison test! There are many choices that fit the bill. (This is assuming you start at $n=1$, you might consider forgetting your singularity point).

answered Apr 23 '13 at 21:11

Coffee_Table

2,909

score 0 · Answer 4 · answered Apr 23 '13 at 21:13

$0^0$ is not so much a point of singularity than a place that has no defined value. Without context, it could be any value.

For example, I am tempted to say that

$$0^0 = \lim_{x \to 0} x^x = \lim_{x \to 0} e^{ x \log{x}} = 1$$

But there are other ways to get there...

Convergence of $\sum_{n=0}^\infty \frac{1}{n^n}$

4 Answers4