How do you determine if the series the converges conditionally, absolutely or diverges given Sigma (cos(npi))/(n+1) from [1,oo)?

1 Answer
Apr 1, 2017

sum_{n=1}^infty{cos(n pi)}/{n+1} converges conditionally.

Explanation:

Since cos(npi)=(-1)^n,

sum_{n=1}^infty{cos(n pi)}/{n+1}=sum_{n=1}^infty{(-1)^n}/{n+1}

Let us see if it is absolutely convergent.

sum_{n=1}^inftyabs{{(-1)^n}/{n+1}}=sum_{n=1}^infty1/{n+1}=1/2+1/3+1/4+cdots,

which is a harmonic series (divergent). So, it is NOT absolutely convergent.

Let us see if it is conditionally convergent.

Since 1/{n+1} is decreasing and lim_{n to infty}1/{n+1}=0, by Alternating Series Test, we know that the series is convergent.

Hence, the series is conditionally convergent.