How do you determine the convergence or divergence of #Sigma ((-1)^(n)n^2)/(n^2+1)# from #[1,oo)#?

Question

Emily Ammerman · Accepted Answer

#sum_(n=1)^oo (-1)^n n^2/(n^2+1)# does not converge

This is an alternating series, so the necessary condition for it to converge is that: #lim a_n = 0# #a_(n+1)/a_n < 1# As: #a_n = n^2/(n^2+1)# we have: #lim_n a_n = lim_n n^2/(n^2+1) = 1# Therefore the series does not converge.

Emma Aldridge · Answer

undefined To determine the convergence or divergence of the series $ \sum_{n=1}^{\infty} \frac{(-1)^{n}n^2}{n^2+1} $ from $ n = 1 $ to $ \infty $, we can use the Alternating Series Test. We need to check if the series satisfies two conditions:

1. The terms $ a_n $ are positive.
2. The terms $ a_n $ are decreasing.

If both conditions are met, then the series converges.

In this case, the terms $ \frac{(-1)^{n}n^2}{n^2+1} $ are always positive because $ n^2 $ and $ n^2+1 $ are positive for all positive integers $ n $.

Now, we need to check if the terms are decreasing. We can do this by considering the absolute value of the terms and showing that they are decreasing.

$$ |a_n| = \frac{n^2}{n^2+1} $$

To show that $ |a_n| $ is decreasing, we can consider $ |a_{n+1}| - |a_n| $:

$$ |a_{n+1}| - |a_n| = \frac{(n+1)^2}{(n+1)^2 + 1} - \frac{n^2}{n^2 + 1} $$

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

$$ = \frac{n^4 + 2n^3 + 2n^2 + 1 - (n^4 + 2n^3 + 3n^2 + 2n + 1)}{(n+1)^2(n^2 + 1)} $$

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

Since $ -n^2 - 2n - 1 $ is always negative for $ n \geq 1 $, and the denominator is always positive, $ |a_{n+1}| - |a_n| $ is negative, indicating that $ |a_n| $ is decreasing.

Since both conditions of the Alternating Series Test are satisfied, the series $ \sum_{n=1}^{\infty} \frac{(-1)^{n}n^2}{n^2+1} $ converges.

Luke Alvarez · Answer

undefined To determine the convergence or divergence of the series $ \sum_{n=1}^{\infty} \frac{(-1)^n n^2}{n^2 + 1} $, you can use the Alternating Series Test.

First, check if the series satisfies the conditions of the Alternating Series Test:
1. The terms of the series alternate in sign: Yes, the terms alternate between positive and negative.
2. The absolute value of the terms decreases as $ n $ increases: We can observe that $ \frac{n^2}{n^2 + 1} $ tends to 1 as $ n $ approaches infinity.

Since the series satisfies both conditions, we can conclude that the series converges.