Determine whether the series converges or diverges.

[Sabricd]

Hello,

I have to determine whether the given series diverges or converges. \sum cos(n*pi)/(n^(3/4)) where n= 1 and goes to infinity.

I tried a couple numbers for n and got:

-1 + 1/(2)^(3/4) - 1/(3)^(3/4)

Hence I came up with the series: \sum((-1)^n)/(n)^(3/4) where n=1 and goes to infinity.

I guess my main question is that now that I have that new representation of the series, why can't I just take the absolute value of the series and say that it is a p series with p< 1 and therefore diverges.

My book took the limit of the series and got 0 and said it converged. Why can't you use the p-series test?


Thank you,

 

[Outlined]

\sum_{n &gt; 0}\frac{(-1)^{n}}{n^{3/4}}

It is an alternating series and therefore it converges. With alternating series you have to use other tests.

see also http://en.wikipedia.org/wiki/Alternating_series_test

 

[Sabricd]

So when you have alternating series you cannot use the p-series test?

 

[micromass]

But the p-series is for series of the form

\sum_{n=1}^\infty{\frac{1}{n^p}}

You have a factor (-1)^n extra.
I guess that you could say that you just took the absolute value of each term. But it is NOT TRUE to say that

\sum{x_n}~\text{diverges}~\text{if}~\sum{|x_n|}~\text{diverges}

if you replace diverges with converges, then it is true however...