Determine whether the infinite series,

Question

sigma(((-1)^(n+1))/n)^2 converges or diverges.

My professor gave these in a problem set after he taught the alternating series test.

Simplying the series we get,

sigma(((-1)^(n+1))/n)^2 =sigma(((-1)^(2n+2))/(n^2) = sigma((1^2)/(n^2) = sigma(1/n^2)

So my question is can we actually apply alternating series test for this problem or we have to use another suitable test?

If alternating series test to be applied my thought was to write the original series as,

sigma(((-1)^(n+1))/n)^2 = sigma(((-1)^(2n+2))/(n^2))= sigma (((-1)^(2n-1)) * (((-1)^2)/(n^2) = sigma(((-1)^(2n-1)) * (1/(n^2)),

Since (-1)^(2n-1)=(-1)^(n-1)

Original series simplifies to sigma(((-1)^(n-1)) * (1/(n^2)),

Taking sigma(a_n)= sigma(1/(n^2)),

We have the original series,

sigma(((-1)^(n-1)) * (1/(n^2))= sigma(((-1)^(n-1))* (a_n)

Since a_n satisfies the conditions of Alternating Series Test, sigma(a_n) converges, thus making the original series convergent.

Or can we simply use the p-series test, since the original series converges to sigma(1/(n^2))?

Thank you!

Answer 1

I guess I made a mistake in the middle of the process where I'm using Alternating Series Test for thus problem by taking,

(-1)^(2n-1)=(-1)^(n-1)

Since this not true for all n, can we use Alternating Series Test for this problem?

Answer 2

since 2n-1 is odd, (-1)^(2n-1) = -1

Answer 3

Can we apply alternating series test for this question?