Chapter 9: Problem 62
For the following infinite series, find the first four terms of the sequence of partial sums. Then make a conjecture about the value of the infinite series. $$1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dots$$
Chapter 9: Problem 62
For the following infinite series, find the first four terms of the sequence of partial sums. Then make a conjecture about the value of the infinite series. $$1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dots$$
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Get started for freeDetermine whether the following series converge absolutely or conditionally, or diverge. $$\sum_{k=1}^{\infty} \frac{(-1)^{k}}{\sqrt{k}}$$
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
Give an argument, similar to that given in the text for the harmonic series, to show that \(\sum_{k=1}^{\infty} \frac{1}{\sqrt{k}}\) diverges.
Determine whether the following series converge absolutely or conditionally, or diverge. $$\sum_{k=1}^{\infty} \frac{(-1)^{k+1} e^{k}}{(k+1) !}$$
Suppose an alternating series \(\sum(-1)^{k} a_{k}\) converges to \(S\) and the sum of the first \(n\) terms of the series is \(S_{n}\) Suppose also that the difference between the magnitudes of consecutive terms decreases with \(k\). It can be shown that for \(n \geq 1,\) $$\left|S-\left[S_{n}+\frac{(-1)^{n+1} a_{n+1}}{2}\right]\right| \leq \frac{1}{2}\left|a_{n+1}-a_{n+2}\right|$$ a. Interpret this inequality and explain why it gives a better approximation to \(S\) than simply using \(S_{n}\) to approximate \(S\). b. For the following series, determine how many terms of the series are needed to approximate its exact value with an error less than \(10^{-6}\) using both \(S_{n}\) and the method explained in part (a). (i) \(\sum_{k=1}^{\infty} \frac{(-1)^{k}}{k}\) (ii) \(\sum_{k=2}^{\infty} \frac{(-1)^{k}}{k \ln k}\) (iii) \(\sum_{k=2}^{\infty} \frac{(-1)^{k}}{\sqrt{k}}\)
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