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

This is an alternating series, so the necessary condition for it to converge is that:

As:

we have:

Therefore the series does not converge.

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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:

- The terms ( a_n ) are positive.
- 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.

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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:

- The terms of the series alternate in sign: Yes, the terms alternate between positive and negative.
- 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.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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