Recent content by Ratio Test =)

\sum_{n=1}^{\infty} a_n \, sin(n) converges whenever {an} is a decreasing sequence that tends to zero. By : "[URL Black"][SIZE="2"]Dirichlet's test[/B][/U][/URL]

\frac{x+1}{3(x-2)^2}=\frac{1}{3} \left( \frac{A}{x-2} + \frac{B}{(x-2)^2} \right) = \frac{A}{3(x-2)} + \frac{B}{3(x-2)^2}

Recall that : \sqrt{1+sin^2(x)}=\sqrt{1+1-cos^2(x)}=\sqrt{2-cos^2(x)} Use t=cos(x). Then do a trigonometric substitution.

Its a riemann sum.

One Question : Who told you this is for my homework ? Did you read the thread's title ?

Ohhh. BTW, Its a famous mistake. :)

Yeah Its wrong. You will face: \int_0^4 \sqrt{ 1 + \frac{9}{4}x } \;\ dx What did you do for it?

This is not an arcsine integral. :redface: Try x=a \,\ tan(\theta).

http://www.wolframalpha.com/input/?i=integrate+sqrt%5B+1+%2B+%5B+%283%2F2%29+x%5E%281%2F2%29+%5D%5E2+%5D+from+x%3D0+to+x%3D4 Its better to post all of your work.

[SIZE="2"]Hello :) Here we go! : [SIZE="2"]1. \int \frac{dx}{x^3+x^2+x+1} [SIZE="2"]2. \int_{\frac{\pi}{2}}^{\pi} \left( sin(x) ln(x) - \frac{cos(x)}{x}\right) dx [SIZE="2"] Do your best :)

Ohhh My bad How in Earth I did not notice this :@

Thanks you. you saved my life ! It is easy for me. It converges by integral test.

No No Sorry! there was a typo! its 2n-2 not 2n-1 in the series =) Sorry Again.

Oh...Thanks But will not affect the convergence right? since the series which obtained by multiplying a convergent series by a constant is still convergent. Right?

nth terms test = Test for divergence. hmmm .. I got it .. Limit comparison test with the divergent series \sum_{n=1}^{\infty} \frac{1}{n+1} The limit of the limit comparison test will give 1/e which is positive and finite number hence, our series diverges. Right?