using integral test for sum of a series

**acosta0809** · July 29th 2009, 11:11 AM

the question is :

Determine whether the integral test can be used to determine the convergence or divergence of the following series. If it can be used, use it to determine the convergence or divergence. If the integral test can not be used, explain why.

$\sum_{n = 0}^{\infty} \frac{\cos(n)}{n}$

what i see is that

$-1\leq\cos(x)\leq1$

so

$\frac{-1}{x}\leq\frac{\cos(x)}{x}\leq\frac{1}{x}$

so because it is not always positive we can not use the integral test.... is this correct?

**DeMath** · July 29th 2009, 11:57 AM

Maybe the amount from one to infinity?

$\sum\limits_{n = {\color{red}{1}}}^\infty {\frac{{\cos n}}{n}}$

**HallsofIvy** · July 29th 2009, 12:21 PM

I don't see how changing the lower index from 0 to 1 helps at all.

cos(2)= -0.41614 which is negative and the integral tests, as acosta0809 suggests, does not apply here.

**acosta0809** · July 29th 2009, 12:34 PM

sorry! yes initial n=1

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