# What is infinity to the power zero

Edited

I have this notation $\lim_{k->\infty} k^ {1/k}$. Is it correct to say that the output is 1, or there is some other result.

P.S: Okay guys made a mistake, sorry.Now please cool down. I am not here for asking silly questions about infinity. Again, sorry.I was in a hurry.

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Which one are you asking, $\sum_{k=1}^\infty k^{1/k}$, as the question says, or $\lim_{k\to\infty} k^{1/k}$, as the topic seems to imply? – JiK Mar 16 '14 at 11:56
First, should the sum be over $k$ instead of $i$? Second, the title of your question is meaningless and should be changed. Third, try to show that $k^{1/k} \geq 1$ for all $k$. – Paul Siegel Mar 16 '14 at 11:57
@Sabyasachi "Indeterminate" is just a fancy word for "meaningless". – Paul Siegel Mar 16 '14 at 12:01
@Sabyasachi "$1^\infty$" is complete nonsense; $e$ is the perfectly well-defined limit of the sequence $(1 + 1/n)^n$ as $n \to \infty$. – Paul Siegel Mar 16 '14 at 12:04
@Sabyasachi Understanding why these expressions are meaningless requires no understanding: $\infty$ is not a number, and cannot be manipulated as such. Would anybody ask whether ${car}^0$ is meaningful? – Paul Siegel Mar 16 '14 at 12:06

$$\lim_{n\to\infty}n^{^\tfrac1n}=\lim_{n\to\infty}\Big(e^{\ln n}\Big)^{^{\tfrac1n}}=\lim_{n\to\infty}e^{^\tfrac{\ln n}n}=e^{^{\displaystyle{\lim_{n\to\infty}}\tfrac{\ln n}n}}=e^{^{\displaystyle{\lim_{t\to\infty}}\dfrac t{e^t}}}=e^0=1.$$

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This is incorrect. 2 is also a valid answer. As is 3, 4... – Jiminion Apr 28 at 14:58

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.

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Can you review my edited question. I am sorry to create a confusion. – motiur Mar 16 '14 at 13:16
@whoknows I did answer the limit It is 1. – Sabyasachi Mar 16 '14 at 13:17
@whoknows $\lim_{k=1}^\infty$ doesn't make sense – Sabyasachi Mar 16 '14 at 13:31
read a book(or article) on limits – Sabyasachi Mar 16 '14 at 13:31
Excuse me, this has never happened.I am not bad at maths. – motiur Mar 16 '14 at 13:33