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  1. AP Calculus
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Explain the Alternating Series Test.

If lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0limn→∞​an​=0 and ana_nan​ is decreasing, then the alternating series ∑(−1)nan\sum (-1)^n a_n∑(−1)nan​ converges.

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Explain the Alternating Series Test.

If lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0limn→∞​an​=0 and ana_nan​ is decreasing, then the alternating series ∑(−1)nan\sum (-1)^n a_n∑(−1)nan​ converges.

Why is it important that ana_nan​ decreases in the Alternating Series Test?

It ensures that the terms are getting smaller in magnitude, allowing the partial sums to converge.

What happens if lim⁡n→∞an≠0\lim_{n \to \infty} a_n \neq 0limn→∞​an​=0 in an alternating series?

The series diverges by the Divergence Test.

Does the Alternating Series Test determine absolute convergence?

No, it only determines conditional convergence. It doesn't tell us if ∑∣an∣\sum |a_n|∑∣an​∣ converges.

What is the significance of cos⁡(nπ)\cos(n\pi)cos(nπ) in alternating series?

cos⁡(nπ)\cos(n\pi)cos(nπ) is equivalent to (−1)n(-1)^n(−1)n, providing the alternating sign for the series.

State the Alternating Series Test.

If an>0a_n > 0an​>0, lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0limn→∞​an​=0, and ana_nan​ is a decreasing sequence, then the alternating series ∑n=1∞(−1)nan\sum_{n=1}^{\infty} (-1)^n a_n∑n=1∞​(−1)nan​ converges.

How to test ∑n=1∞(−1)nn\sum_{n=1}^{\infty} \frac{(-1)^n}{n}∑n=1∞​n(−1)n​ for convergence?

  1. Identify an=1na_n = \frac{1}{n}an​=n1​. 2. Check lim⁡n→∞1n=0\lim_{n \to \infty} \frac{1}{n} = 0limn→∞​n1​=0. 3. Verify 1n>1n+1\frac{1}{n} > \frac{1}{n+1}n1​>n+11​. Since both conditions are met, the series converges.

How to test ∑n=1∞(−1)nnn+1\sum_{n=1}^{\infty} \frac{(-1)^n n}{n+1}∑n=1∞​n+1(−1)nn​ for convergence?

  1. Identify an=nn+1a_n = \frac{n}{n+1}an​=n+1n​. 2. Check lim⁡n→∞nn+1=1≠0\lim_{n \to \infty} \frac{n}{n+1} = 1 \neq 0limn→∞​n+1n​=1=0. Since the limit is not 0, the series diverges.

How to test ∑n=1∞(−1)nln⁡(n)\sum_{n=1}^{\infty} \frac{(-1)^n}{\ln(n)}∑n=1∞​ln(n)(−1)n​ for convergence?

  1. Identify an=1ln⁡(n)a_n = \frac{1}{\ln(n)}an​=ln(n)1​. 2. Check lim⁡n→∞1ln⁡(n)=0\lim_{n \to \infty} \frac{1}{\ln(n)} = 0limn→∞​ln(n)1​=0. 3. Verify 1ln⁡(n)>1ln⁡(n+1)\frac{1}{\ln(n)} > \frac{1}{\ln(n+1)}ln(n)1​>ln(n+1)1​. Since both conditions are met, the series converges.

How to test ∑n=1∞(−1)nn22n\sum_{n=1}^{\infty} \frac{(-1)^n n^2}{2^n}∑n=1∞​2n(−1)nn2​ for convergence?

  1. Identify an=n22na_n = \frac{n^2}{2^n}an​=2nn2​. 2. Check lim⁡n→∞n22n=0\lim_{n \to \infty} \frac{n^2}{2^n} = 0limn→∞​2nn2​=0 (using L'Hopital's rule). 3. Verify n22n>(n+1)22n+1\frac{n^2}{2^n} > \frac{(n+1)^2}{2^{n+1}}2nn2​>2n+1(n+1)2​ (true for large n). Since both conditions are met, the series converges.