Chapter 1: Problem 5
Test the following series for convergence. $$\sum_{n=2}^{\infty} \frac{(-1)^{n}}{\ln n}$$
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chapter 1: Problem 5
Test the following series for convergence. $$\sum_{n=2}^{\infty} \frac{(-1)^{n}}{\ln n}$$
These are the key concepts you need to understand to accurately answer the question.
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Get started for freeTest the following series for convergence or divergence. Decide for yourself which test is easiest to use, but don't forget the preliminary test. Use the facts stated above when they apply. $$\sum_{n=1}^{\infty} \frac{n^{n}}{n !}$$
Test for convergence: $$\sum_{n=1}^{\infty} \frac{2^{n}}{n !}$$
Test for convergence: $$\sum_{n=2}^{\infty} \frac{1}{n \ln \left(n^{3}\right)}$$
(a) It is clear that you (or your computer) can't find the sum of an infinite series just by adding up the terms one by one. For example, to get \(\zeta(1.1)=\) \(\sum_{n=1}^{\infty} 1 / n^{1.1}(\text { see Problem } 15.22)\) with error \(<0.005\) takes about \(10^{33}\) terms. To see a simple alternative (for a series of positive decreasing terms) look at Figures 6.1 and 6.2. Show that when you have summed \(N\) terms, the sum \(R_{N}\) of the rest of the series is between \(I_{N}=\int_{N}^{\infty} a_{n} d n\) and \(I_{N+1}=\int_{N+1}^{\infty} a_{n} d n\) (b) Find the integrals in (a) for the \(\zeta(1.1)\) series and verify the claimed number of terms needed for error \(<0.005 .\) Hint: Find \(N\) such that \(I_{N}=0.005 .\) Also find upper and lower bounds for \(\zeta(1.1)\) by computing \(\sum_{n=1}^{N} 1 / n^{1.1}+\int_{N}^{\infty} n^{-1.1} d n\) and \(\sum_{n=1}^{N} 1 / n^{1.1}+\int_{N+1}^{\infty} n^{-1.1} d n\) where \(N\) is far less than \(10^{33} .\) Hint: You want the difference between the upper and lower limits to be about \(0.005 ;\) find \(\mathrm{N}\) so that term \(a_{N}=0.005\).
Find the interval of convergence of each of the following power series; be sure to investigate the endpoints of the interval in each case. $$\sum_{n=1}^{\infty} \frac{(x-2)^{n}}{3^{n}}$$
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