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 $\displaystyle \displaystyle \begin{align*} \lim_{x \to \infty}x^x = \infty \end{align*}$.
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.
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.
$\displaystyle \displaystyle \begin{align*} \lim_{x \to \infty}\left|\frac{\frac{1}{(x + 1)^{x + 1}}}{\frac{1}{x^x}}\right| &= \lim_{x \to \infty}\left|\frac{x^x}{(x + 1)^{x + 1}}\right| \\ &= \lim_{x \to \infty}\left|\frac{x^x}{(x + 1)(x + 1)^x}\right| \\ &=\lim_{x \to \infty}\left|\frac{1}{x + 1}\left(\frac{x}{x + 1}\right)^x\right| \\ &= \lim_{x \to \infty}\left|\frac{1}{x + 1}\right| \lim_{x \to \infty}\left|\left(\frac{x}{x + 1}\right)^x\right| \\ &= 0 \lim_{x \to \infty}\left|e^{\ln{\left[\left(\frac{x}{x + 1}\right)^x\right]}}\right| \\ &= 0 \lim_{ x \to \infty}\left|e^{x\ln{\left(\frac{x}{x + 1}\right)}}\right| \\ &= 0\lim_{x \to \infty}\left| e^{\frac{\ln{\left(1 - \frac{1}{x + 1}\right)}}{\frac{1}{x}}} \right| \\ &= 0\left|e^{\lim_{x \to \infty}\left[\frac{\ln{\left(1 - \frac{1}{x + 1}\right)}}{\frac{1}{x}}\right]}\right| \\ &= 0\left|e^{\lim_{x \to \infty}\left(\frac{\frac{1}{x^2 + x}}{-\frac{1}{x^2}}\right)}\right| \textrm{ by L'Hospital's Rule}\end{align*}$
$\displaystyle \displaystyle \begin{align*} &= 0\left|e^{\lim_{x \to \infty}\left(-\frac{x^2}{x^2 + x}\right)}\right| \\ &= 0\left|e^{\lim_{x \to \infty}\left(-\frac{1}{1 + \frac{1}{x}}\right)}\right| \\ &= 0\left|e^{-1}\right| \\ &= 0 \end{align*}$
Since this ratio is less than 1, the series is convergent.