Question

Alternating infinite series

Answer 1

Determine the convergence or divergence of the series.
\[\sum_{n=1}^{\infty}\frac{(-1)^nn^2}{n^2+1}\]

Answer 2

I've tried the ratio test (and it's inconclusive with L=1); am I allowed to use the nth term test to show that the series diverges?

Answer 3

It's clearly divergent because ratio test would suggest that the the series does not have a ratio within the interval of convergence and that is noticed in the very first between the second and the first terms. Also, we notice that the denominator will always be positive as well as the n^2 in the numerator, hence the only time this sequence is negative is when (-1)^n is negative. This means that term in the sum changes sign, which also suggests that the series is divergent.

@inkyvoyd

Answer 4

@genius12 , the ratio test results in a limit of L=1, which is indeterminant. For alternating series tests I am given one other test that doesn't return determinant results -
For alternating series, the series (-1)^n b converges if
1. The limit of b_n as n approaches infinity is zero
2. B is a decreasing sequence.
This obviously doesn't apply to the series, but I can't assume the converse is necessarily true because this isn't and iff statement.

Answer 5

Since I have to "justify" my answer, I cannot simply say that with the picture of the plot of the sequence it appears to diverge.

Answer 6

Simply note that the series does not converge to zero, and alternates as it goes to infinity. Hence it has no true summation.

So, using the nth term test, you can note that this is the case.

Answer 7

what @LolWolf said
the limit does not exist (that is what the nth term test says) if
\[\lim_{n\to \infty}a_n\] does not exist, then neither does the sum

Answer 8

Okay - tanks guys