What is infinity to the power zero I have this notation:
$$\lim_{k->\infty} k^ {1/k}$$
Is it correct to say that the output is 1, or is there some other result?
 A: $$\lim_{n\to\infty}n^{^\tfrac1n}=\lim_{n\to\infty}\Big(e^{\ln n}\Big)^{^{\tfrac1n}}=\lim_{n\to\infty}e^{^\tfrac{\ln n}n}=e^{^{\displaystyle{\lim_{n\to\infty}}\tfrac{\ln n}n}}=e^{^{\displaystyle{\lim_{t\to\infty}}\dfrac t{e^t}}}=e^0=1.$$
A: There is no universal value for $\infty^0$. It is indeterminate, and the value depends on how you are getting the $\infty$ and the $0$. Some other indeterminate forms are $0^0, 1^\infty, \infty\times0,\frac00, 1$. I might have missed a few. 
For example consider the function $f_1(n)=(1+\frac1n)^n$. At $\infty$ it is of the form $1^\infty$, but 
$$\lim_{n\to\infty} f(n)=e\approx2.718\cdots$$
Now consider $f_2(n) = (1+\frac2n)^n$. At $\infty$ this is also of the form $1^\infty$ but the limit is,
$$\lim_{n\to\infty} f(n)=e^2\approx7.389\cdots$$
As for your actual question, $\sum_{i=1}^\infty k^{1/k}$ is infinite since you are adding a non-zero constant $(k^{1/k})$ to itself an infinite number of times. 
If we change the summation to $\sum_{i=1}^\infty i^{1/i}$, this is still infinite as although the terms are not constant, each of them is greater than $1$ and so the series is greater than $1+1+1+\cdots$, therefore infinite. 
As to why each term is greater than $1$, the function $f(x)=x^{1/x}$ is monotonically decreasing $\forall x\gt e$, and each term must be greater than the limit of the function at $\infty$, and this limit
$$\lim_{n\to\infty} n^{1/n} = 1$$
As to why this limit is $1$,(and why the other limits are $e$ and $e^2$ respectively) this is a topic too broad for the scope of this answer. I suggest you read the article linked above, and also read a good book on calculus, in case you are willing to self-study.
