Why does this infinite series diverge?

[kmacinto]

Homework Statement

Why does this series diverge?

Homework Equations

$\sum_{n}^{\infty }\frac{-1^{(2n+2)}}{n+1}$

The Attempt at a Solution

I must be missing a rule with the -1 sign.

My logic is that for all n, the numerator = 1 since anything to the power of 2 is even and adding the 2 doesn't change the sign either. So if the numerator is always 1 and the denominator grows without bound so by the alternating series test since, L = 0 and the series is decreasing, the series should converge... My math book and maple says different though :(

 

[SammyS]

Homework Statement

Why does this series diverge?

Homework Equations

$\sum_{n}^{\infty }\frac{-1^{(2n+2)}}{n+1}$

The Attempt at a Solution

I must be missing a rule with the -1 sign.

My logic is that for all n, the numerator = 1 since anything to the power of 2 is even and adding the 2 doesn't change the sign either. So if the numerator is always 1 and the denominator grows without bound so by the alternating series test since, L = 0 and the series is decreasing, the series should converge... My math book and maple says different though :(

As you pointed out, it's not an alternating series.

For clarity, put parentheses around the -1. _ _ _ _ (-1)

$\displaystyle \sum_{n}^{\infty }\frac{(-1)^{(2n+2)}}{n+1}$

 

[shaon0]

Homework Statement

Why does this series diverge?

Homework Equations

$\sum_{n}^{\infty }\frac{-1^{(2n+2)}}{n+1}$

The Attempt at a Solution

I must be missing a rule with the -1 sign.

My logic is that for all n, the numerator = 1 since anything to the power of 2 is even and adding the 2 doesn't change the sign either. So if the numerator is always 1 and the denominator grows without bound so by the alternating series test since, L = 0 and the series is decreasing, the series should converge... My math book and maple says different though :(

S=Ʃ(-1)^2k+2/(k+1)
Try a simpler case of the above. Try the other tests. Maybe the integral test esp. something looking like a log fn differentiated.

 

[kmacinto]

I think I have it... If the numerator is positive for all n, I can just use the direct comparison test with B sub N = 1/n which is a divergent p-series and since B sub N is less than A sub N, A sub N also diverges... right?

 

[SammyS]

That sounds good to me !