Convergence of an infinite series

edoz
Messages
2
Reaction score
0

Homework Statement


http://img840.imageshack.us/img840/3609/unleddn.png

note that by log(n), i really mean NATURAL log of n

Homework Equations


it's convergent, but I can't figure out which test to use

The Attempt at a Solution


there is no term to the nth power, so ratio test is useless; root test is useless too; comparison test would seem to be the best option, but I can't figure out how to compare when I have a natural log in the expression... limit comparison test takes me nowhere either.
 
Last edited by a moderator:
Physics news on Phys.org
I think that the log(n) is just there to throw you off. \log(n) < n^{1/2} asymptotically right. Do you have a theorem showing that \sum \frac{1}{n^{3/2}} is convergent?
 
yes, p-series are convergen if p>1;

now i get it, I just need to compare it with the series you just shown, since with the (n+1) the original series will always be smaller than the convergent series you shown, therefore convergent too.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

Similar threads

Replies
3
Views
1K
Replies
1
Views
1K
Replies
5
Views
2K
Replies
6
Views
2K
Replies
4
Views
2K
Replies
14
Views
2K
Replies
4
Views
1K
Back
Top