How do you test the alternating series #Sigma (1)^n(sqrt(n+1)sqrtn)# from n is #[1,oo)# for convergence?
See below.
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To test the convergence of the alternating series ( \sum_{n=1}^{\infty} (1)^n(\sqrt{n+1}\sqrt{n}) ), we can use the Alternating Series Test.

Check for the conditions of the Alternating Series Test:
 The terms ( a_n = \sqrt{n+1}  \sqrt{n} ) are decreasing.
 The limit of the terms ( a_n ) as ( n ) approaches infinity is 0.

To verify the first condition, we can observe that as ( n ) increases, ( \sqrt{n+1}  \sqrt{n} ) will approach 0, indicating decreasing terms.

To verify the second condition, we calculate the limit of ( \sqrt{n+1}  \sqrt{n} ) as ( n ) approaches infinity. By rationalizing the expression, we get: [ \lim_{n \to \infty} (\sqrt{n+1}  \sqrt{n}) = \lim_{n \to \infty} \frac{(\sqrt{n+1}  \sqrt{n})(\sqrt{n+1} + \sqrt{n})}{\sqrt{n+1} + \sqrt{n}} ] Simplifying, we get: [ \lim_{n \to \infty} (\sqrt{n+1}  \sqrt{n}) = \lim_{n \to \infty} \frac{(n+1)  n}{\sqrt{n+1} + \sqrt{n}} = \lim_{n \to \infty} \frac{1}{\sqrt{n+1} + \sqrt{n}} = 0 ]

Since both conditions of the Alternating Series Test are satisfied, we can conclude that the alternating series ( \sum_{n=1}^{\infty} (1)^n(\sqrt{n+1}\sqrt{n}) ) converges.
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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