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Chapter 7: Problem 93

Find the sum of the series \(\sum_{n=2}^{\infty} \ln \left(1-\frac{1}{n^{2}}\right)\).

Short Answer

Expert verified
The sum of the series \(\sum_{n=2}^{\infty} \ln(1-\frac{1}{n^{2}})\) is \(\ln(4)\)

Step by step solution

01

Identify the series

The series given is \(\sum_{n=2}^{\infty} \ln(1-\frac{1}{n^{2}})\)
02

Rewrite the series

Redefine the series using logarithm properties. As \(\ln(ax) = \ln(a) + \ln(x)\), and \(\ln(1-\frac{1}{n^{2}}) = \ln( \frac{n^{2}-1}{n^{2}} ) = \ln(n^{2}) - \ln(n^{2}-1)\). So the series now becomes \(\sum_{n=2}^{\infty}[\ln(n^{2}) - \ln(n^{2}-1)]\).
03

Split the series

Split the series into two, with the general terms \(\ln(n^{2})\) and \(- \ln(n^{2}-1)\). So, the series can be written as \(\sum_{n=2}^{\infty}\ln(n^{2}) - \sum_{n=2}^{\infty}\ln(n^{2}-1)\)
04

Identify the telescoping series

A telescoping series is formed in \(\sum_{n=2}^{\infty}\ln(n^{2}) - \sum_{n=2}^{\infty}\ln(n^{2}-1)\). Here, each term in \(\sum_{n=2}^{\infty}\ln(n^{2}-1)\) cancels out with the corresponding term in \(\sum_{n=2}^{\infty}\ln(n^{2})\). Specifically, the term in the first series when n=k cancels out with the term in the second series when n=k+1.
05

Find the sum of the series

After the cancellation in the telescoping series, the remaining terms are \(\ln(4) - \ln(1) = \ln(4)\). As log to the base e of 1 is 0, the sum of the series is \(\ln(4)\).

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Most popular questions from this chapter

Find all values of \(x\) for which the series converges. For these values of \(x,\) write the sum of the series as a function of \(x\). $$ \sum_{n=1}^{\infty}\left(\frac{x^{2}}{x^{2}+4}\right)^{n} $$

In an experiment, three people toss a fair coin one at a time until one of them tosses a head. Determine, for each person, the probability that he or she tosses the first head. Verify that the sum of the three probabilities is 1 .

Find the sum of the convergent series. $$ \sum_{n=1}^{\infty} \frac{1}{(2 n+1)(2 n+3)} $$

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(a) write the repeating decimal as a geometric series and (b) write its sum as the ratio of two integers $$ 0 . \overline{81} $$
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