Proving the divergence of arcsin(1/n)

  • Thread starter Thread starter mybluesock
  • Start date Start date
  • Tags Tags
    Divergence
mybluesock
Messages
1
Reaction score
0

Homework Statement


Is \sum(-1)^(n-1)*arcsin(1/n) absolutely convergent, conditionally convergent, or divergent?


2. The attempt at a solution

The original function is alternating, so by the alternating series test, the function is convergent, because 0 < arcsin(1/(n+1)) <arcsin(1/n), and the limit of arcsin(1/n)=0.
So that rules out divergent. To determine whether the series is absolutely or conditionally convergent, you test the convergence of the absolute value of the series, which would be \sum arcsin(1/n). However, I'm not sure what test to use now. Should I use the comparison test, and if so, what should I compare it to?
 
Physics news on Phys.org
i would attempt to use a comparison test...

considering the taylor series of arcsin will also be useful in showing the comparison is valid
 
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
6
Views
2K
Replies
1
Views
1K
Replies
2
Views
2K
Replies
4
Views
1K
Replies
14
Views
2K
Back
Top