Determine whether the infinite series,
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!
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?
since 2n-1 is odd, (-1)^(2n-1) = -1
Can we apply alternating series test for this question?
To determine whether the infinite series
Σ (((-1)^(n+1))/n)^2
converges or diverges, we can use the alternating series test or the p-series test.
First, let's simplify the series:
Σ (((-1)^(n+1))/n)^2 = Σ (((-1)^(2n+2))/(n^2)) = Σ ((1^2)/(n^2)) = Σ (1/n^2)
Now, let's consider using the alternating series test. We can rewrite the original series as:
Σ (((-1)^(n+1))/n)^2 = Σ (((-1)^(2n-1)) * ((1^2)/(n^2)))
Since (-1)^(2n-1) = (-1)^(n-1), we can simplify further:
Σ (((-1)^(n-1)) * (1/(n^2)) = Σ ((-1)^(n-1)) * (a_n)
Here, a_n = 1/(n^2) satisfies the conditions of the alternating series test, as a_n is decreasing, approaches zero as n approaches infinity, and is positive for all n.
Therefore, we can conclude that the series Σ (((-1)^(n+1))/n)^2 converges, since it satisfies the conditions of the alternating series test.
Alternatively, we can also apply the p-series test. The p-series test states that if we have a series Σ (1/n^p), where p > 1, the series converges. In this case, our original series Σ (1/n^2) has p = 2, which is greater than 1. Hence, the series converges.
Both methods lead us to the same conclusion that the infinite series
Σ (((-1)^(n+1))/n)^2
converges.