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Chapter 9: Problem 44
Use the properties of infinite series to evaluate the following series. $$\sum_{k=2}^{\infty} 3 e^{-k}$$
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Determine whether the following series converge absolutely or conditionally, or diverge. $$\sum_{k=2}^{\infty} \frac{(-1)^{k}}{\ln k}$$
Showing that \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6}
\operatorname{In} 1734,\) Leonhard Euler informally
proved that \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6} .\) An
elegant proof is outlined here that uses the inequality
$$
\cot ^{2} x<\frac{1}{x^{2}}<1+\cot ^{2} x\left(\text { provided that }
0
Convergence parameter Find the values of the parameter \(p>0\) for which the following series converge. $$\sum_{k=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots(2 k-1)}{p^{k} k !}$$
Consider the sequence \(\left\\{F_{n}\right\\}\) defined by $$F_{n}=\sum_{k=1}^{\infty} \frac{1}{k(k+n)^{\prime}}$$ for \(n=0,1,2, \ldots . .\) When \(n=0,\) the series is a \(p\) -series, and we have \(F_{0}=\pi^{2} / 6\) (Exercises 65 and 66 ). for \(n=0,1,2, \ldots . .\) When \(n=0,\) the series is a \(p\) -series, and we have \(F_{0}=\pi^{2} / 6\) (Exercises 65 and 66 ).
Find a series that a. converges faster than \(\sum \frac{1}{k^{2}}\) but slower than \(\sum \frac{1}{k^{3}}\) b. diverges faster than \(\sum \frac{1}{k}\) but slower than \(\sum \frac{1}{\sqrt{k}}\) c. converges faster than \(\sum \frac{1}{k \ln ^{2} k}\) but slower than \(\sum \frac{1}{k^{2}}\)
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