Series and convergence/divergence

  • Thread starter Thread starter Niles
  • Start date Start date
  • Tags Tags
    Series
Niles
Messages
1,834
Reaction score
0

Homework Statement


Determine whether the series converges:

<br /> \sum\limits_{n = 2}^{\inf } {\frac{{( - 1)^n (n^2 + 1)^{1/2} }}{{n\ln (n)}}} <br />


The Attempt at a Solution



Which test must I use? I thought of using the integral test, but it seems a little too hard. Are there other possibilities?
 
Physics news on Phys.org
Try either the ratio or root test. Neither would be very easy to calculate, but...
 
Well, the (-1)^n should give you an easy test...

BTW: \infty is \infty
 
Should I just take:

<br /> \sum\limits_{n = 2}^{\inf } {\frac{{( - 1)^n (n^2 + 1)^{1/2} }}{{n\ln (n)}}}^{(1/n)} <br /> and take the limit of this?
 
Its an alternating series, what about the Leibinitz test?
 
I looked at the Leinitz test - on wikipedia they write about the first condition:

"If the sequence a_n converges to 0, and .." - does this mean the limit of a_n for n -> infinity?
 
Last edited:
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

Back
Top