Thursday

March 5, 2015

March 5, 2015

Posted by **kelly** on Tuesday, April 17, 2007 at 6:03pm.

Sin(n) is always between -1 and 1. So, you have:

-1/n^2 < a(n) < 1/n^2

Both the lower bound and the upper bound converge to zero, so the sequence converges to zero.

**Answer this Question**

**Related Questions**

Calculus - Determine whether the sequence A(n)=nsinn/n^3 converges or diverges. ...

Calculus II - Determine whether the sequence converges or diverges. If it ...

calculus - 1. integral -oo, oo [(2x)/(x^2+1)^2] dx 2. integral 0, pi/2 cot(theta...

Calc - Does 1/ln(x+1) converge or diverge? I've tried the nth term test, limit ...

CALCULUS - For the sequence below, determine whether they converge or not, and ...

Calculus - The problem with these two questions is that I cannot determine the a...

Calculus - integral -oo, oo [(2x)/(x^2+1)^2] dx (a) state why the integral is ...

calc 2 - Determine whether the integral converges or diverges. Find the value of...

calculus - Determine whether the given series converges or diverges, and find ...

Calculus: Limits - ++++++++|||||||||Can you please check whether or not my ...