# Convergence of $\sum_{n=0}^\infty \frac{1}{n^n}$ [duplicate]

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.

-

## marked as duplicate by Eric NaslundApr 23 '13 at 21:21

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 math.stackexchange.com/questions/11150/zero-to-zero-power and I have closed it. – Eric Naslund Apr 23 '13 at 21:22

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 $n=1$.

-

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

-

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).

-

$0^0$ 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...

-