Originally Posted by
Slyprince You can use the ratio test
$\displaystyle \lim_{n\rightarrow \infty}\left | \frac{n+1}{\left (\ln (n+1) \right )^{n+1}}\cdot\frac{(\ln n)^{n}}{n} \right |$
Expand the n+1 exponent
$\displaystyle \lim_{n\rightarrow \infty} \frac{n+1}{\left (\ln (n+1) \right )^{n}\cdot \ln (n+1)}\cdot\frac{(\ln n)^{n}}{n}$
Divide stuff so that using limit laws so it is easier to take the limit
$\displaystyle \lim_{n\rightarrow \infty }\frac{n+1}{n}\cdot \lim_{n\rightarrow \infty }\frac{\left (\ln n \right )^n}{\left ( \ln (n+1) \right )^n}\cdot \lim_{n\rightarrow \infty }\frac{1}{\ln (n+1)}$
The first limit goes to 1 and the third limit goes to zero, but the second needs more work so raise the whole thing to the shared exponent
$\displaystyle \lim_{n\rightarrow \infty }\left (\frac{\ln n}{\ln (n+1)} \right )^{n}\cdot 0$
you can move the limit inside because of the power limit law
$\displaystyle \left (\lim_{n\rightarrow \infty }\frac{\ln n}{\ln (n+1)} \right )^{n}\cdot 0$
Uh? What is that rule? And anyway no, you cannot move the limit "inside" when ALSO the power depends on the variable that is in the limit! If you
could then you'd get $\displaystyle e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n=\l eft(\lim_{n\to\infty}\left(1+\frac{1}{n}\right)\ri ght)^n=(1)^n=1$ , which of course is wrong.
In this case the limit certainly is 1, but the way to get there is wrong.
Tonio
now you can use l'hospital's rule on the inside
$\displaystyle \left (\lim_{n\rightarrow \infty }\frac{\frac{1}{n}}{\frac{1}{n+1}} \right )^{n}\cdot 0$
with inverting and flipping you get this...
$\displaystyle \left (\lim_{n\rightarrow \infty }\frac{n+1}{n} \right )^{n}\cdot 0$
the limit goes to 1 and multiplied by zero gives zero which is less than 1 so by the ratio test...
math]\sum_{n=2}^{\infty }\frac{n}{\left (\ln n \right )^{n}}[/tex] converges
whoo, that was a mouthful