# Thread: Simple problem involving comparison test for series...

1. ## Simple problem involving comparison test for series...

I have a simple question involving the comparison test for series. I keep getting that it diverges, but the book says it diverges and I'm not getting why that is. The problem asks whether the series from n=1 to infinity converges. The expression is [cos(n*pi)]/[n^(3/4)]. My reasoning is that the numerator, the cos(n*pi) is always equal to or greater to 1, so it could be compared to 1/n^(3/4), which is a p-series with p=3/4, so it diverges. Therefore, the original expression diverges as well since it is smaller than the one it was compared to. But that's not right. What should I compare it to? I now notice that the cos(n*pi) can be negative, but how does that affect the problem? Please help; thanks.

2. Originally Posted by virtuoso735
I have a simple question involving the comparison test for series. I keep getting that it diverges, but the book says it diverges and I'm not getting why that is. The problem asks whether the series from n=1 to infinity converges. The expression is [cos(n*pi)]/[n^(3/4)]. My reasoning is that the numerator, the cos(n*pi) is always equal to or greater to 1, so it could be compared to 1/n^(3/4), which is a p-series with p=3/4, so it diverges. Therefore, the original expression diverges as well since it is smaller than the one it was compared to. But that's not right. What should I compare it to? I now notice that the cos(n*pi) can be negative, but how does that affect the problem? Please help; thanks.

$\sum_{n = 1}^\infty \frac{ \cos n \pi}{n^{3/4}} = \sum_{n = 1}^\infty \frac{ (-1)^n}{n^{3/4}}$

3. Originally Posted by virtuoso735
I have a simple question involving the comparison test for series. I keep getting that it diverges, but the book says it diverges and I'm not getting why that is. The problem asks whether the series from n=1 to infinity converges. The expression is [cos(n*pi)]/[n^(3/4)]. My reasoning is that the numerator, the cos(n*pi) is always equal to or greater to 1, so it could be compared to 1/n^(3/4), which is a p-series with p=3/4, so it diverges. Therefore, the original expression diverges as well since it is smaller than the one it was compared to. But that's not right. What should I compare it to? I now notice that the cos(n*pi) can be negative, but how does that affect the problem? Please help; thanks.
Notice for n=1,2,3,...

$\cos(n\pi)=-1,1,-1,1.....$

$\sum_{n=1}^{\infty}\frac{\cos(n\pi)}{n^{\frac{3}{4 }}} =\sum_{n=1}^{\infty}\frac{(-1)^n)}{n^{\frac{3}{4}}}$

This series converges by the Alternating series test becuase

$a_n=\frac{1}{n^{3/4}}, \lim_{n \to \infty}a_n=0$

4. Thanks for the replies; they were helpful. However, in this part of the book, we haven't went over the alternating series test yet, so how would this be solved without referring to that test?