# Thread: x to power x convergence test

1. ## x to power x convergence test

How do i prove the positive term sequence x^x converges using Cauchy integral test ?

I am not able to figure out how to integrate x^x ?

Any ideas ?

2. ## Re: x to power x convergence test

Originally Posted by pratique21
How do i prove the positive term sequence x^x converges using Cauchy integral test ?

I am not able to figure out how to integrate x^x ?

Any ideas ?
Why should you need the integral test? Since it's a sequence, just evaluate \displaystyle \begin{align*} \lim_{x \to \infty}x^x = \infty \end{align*}.

Obviously the sequence doesn't converge, and if it was a series and the terms don't go to 0, the series can't possibly converge either.

3. ## Re: x to power x convergence test

Just what happens when you hold back info !
The sequence I need to check for convergence is $\frac{1}{x^x}$ so i need a way to integrate it. And this one converges
But how do i integrate it ?

4. ## Re: x to power x convergence test

Again, why should you need to integrate it? It's pretty obvious that it goes to \displaystyle \begin{align*} \frac{1}{\infty} = 0 \end{align*}.

5. ## Re: x to power x convergence test

Because that is necessary but not sufficient to prove convergence.

6. ## Re: x to power x convergence test

Originally Posted by pratique21
Because that is necessary but not sufficient to prove convergence.
If it is a SEQUENCE then it is enough to show convergence by evaluating the limit of the terms to a number.

If it is a SERIES then one must first show that the terms go to 0, and then apply some other test.

Which is it?

7. ## Re: x to power x convergence test

umm...does series mean having a summation ?
If yes, then it's a series.

8. ## Re: x to power x convergence test

Originally Posted by pratique21
umm...does series mean having a summation ?
If yes, then it's a series.
A sequence is a list of numbers following a pattern. A series is the sum of the numbers in the sequence.

To show the series is convergent, you can use the ratio test.

\displaystyle \begin{align*} \lim_{x \to \infty}\left|\frac{\frac{1}{(x + 1)^{x + 1}}}{\frac{1}{x^x}}\right| &= \lim_{x \to \infty}\left|\frac{x^x}{(x + 1)^{x + 1}}\right| \\ &= \lim_{x \to \infty}\left|\frac{x^x}{(x + 1)(x + 1)^x}\right| \\ &=\lim_{x \to \infty}\left|\frac{1}{x + 1}\left(\frac{x}{x + 1}\right)^x\right| \\ &= \lim_{x \to \infty}\left|\frac{1}{x + 1}\right| \lim_{x \to \infty}\left|\left(\frac{x}{x + 1}\right)^x\right| \\ &= 0 \lim_{x \to \infty}\left|e^{\ln{\left[\left(\frac{x}{x + 1}\right)^x\right]}}\right| \\ &= 0 \lim_{ x \to \infty}\left|e^{x\ln{\left(\frac{x}{x + 1}\right)}}\right| \\ &= 0\lim_{x \to \infty}\left| e^{\frac{\ln{\left(1 - \frac{1}{x + 1}\right)}}{\frac{1}{x}}} \right| \\ &= 0\left|e^{\lim_{x \to \infty}\left[\frac{\ln{\left(1 - \frac{1}{x + 1}\right)}}{\frac{1}{x}}\right]}\right| \\ &= 0\left|e^{\lim_{x \to \infty}\left(\frac{\frac{1}{x^2 + x}}{-\frac{1}{x^2}}\right)}\right| \textrm{ by L'Hospital's Rule}\end{align*}

\displaystyle \begin{align*} &= 0\left|e^{\lim_{x \to \infty}\left(-\frac{x^2}{x^2 + x}\right)}\right| \\ &= 0\left|e^{\lim_{x \to \infty}\left(-\frac{1}{1 + \frac{1}{x}}\right)}\right| \\ &= 0\left|e^{-1}\right| \\ &= 0 \end{align*}

Since this ratio is less than 1, the series is convergent.