Convergence of series log(1-1/n^2)

In summary, the sum of the series ##\sum\limits_{n=2}^{\infty}\ln\left(1-\dfrac{1}{n^2}\right)## can be found by factoring the terms and working with the resulting product. This leads to a proof that the sum is equal to ##-\ln 2##, which can be further simplified to ##\ln \frac{1}{2}##. The series can also be seen as a sum of two telescoping series, each with a product of terms that can be simplified to 1, leading to the same result.
  • #1
Felipe Lincoln
Gold Member
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Homework Statement


Find the sum of ##\sum\limits_{n=2}^{\infty}\ln\left(1-\dfrac{1}{n^2}\right) ##

Homework Equations


No one.

The Attempt at a Solution


At first I though it as a telescopic serie:
##\sum\limits_{n=2}^{\infty}\ln\left(1-\dfrac{1}{n^2}\right) =\ln\left(\dfrac{3}{4}\right) + \ln\left(\dfrac{8}{9}\right) + \ln\left(\dfrac{15}{16}\right) + \dots < 0##
But it doesn't look like so.
we know from it's terms that ##s_n## is in the interval of ##(L, \ln\left(3/4\right)]##
 
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  • #2
Felipe Lincoln said:

Homework Statement


Find the sum of ##\sum\limits_{n=2}^{\infty}\ln\left(1-\dfrac{1}{n^2}\right) ##

Homework Equations


No one.

The Attempt at a Solution


At first I though it as a telescopic serie:
##\sum\limits_{n=2}^{\infty}\ln\left(1-\dfrac{1}{n^2}\right) =\ln\left(\dfrac{3}{4}\right) + \ln\left(\dfrac{8}{9}\right) + \ln\left(\dfrac{15}{16}\right) + \dots < 0##
But it doesn't look like so.
we know from it's terms that ##s_n## is in the interval of ##(L, \ln\left(3/4\right)]##
You are right about its being a telescoping series. How can you factor ##1-\frac 1 {n^2}##?
 
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  • #3
Using basic property of logarithms you can find that it is equal to
$$\ln\prod_{n=2}^{\infty} \frac{(n-1)(n+1)}{n^2}$$

So you can work with that product instead and prove that the product is equal to ##\frac{1}{2}## so the limit is ##\ln\frac{1}{2}=-\ln2##

I know that usually is not a good idea to work with products but in this case it works..

Now that I see it again, this product is the product of two "telescopic" products ##\prod\frac{n-1}{n},\prod\frac{n+1}{n}##, so I guess the original series must be a sum of two telescopic series.
 
Last edited:

FAQ: Convergence of series log(1-1/n^2)

1. What is the formula for the convergence of series log(1-1/n^2)?

The formula for the convergence of series log(1-1/n^2) is ln(1-1/n^2) = -∑n=1 to ∞ (1/n^2).

2. How do you determine if the series log(1-1/n^2) is convergent or divergent?

The series log(1-1/n^2) is convergent if the limit of the terms as n approaches infinity is equal to zero. In other words, if ∑n=1 to ∞ (1/n^2) is less than infinity, then the series is convergent.

3. What is the interval of convergence for the series log(1-1/n^2)?

The interval of convergence for the series log(1-1/n^2) is (0, ∞), meaning that the series will converge for all values of n greater than zero.

4. Can the convergence of series log(1-1/n^2) be proven using the ratio test?

Yes, the convergence of series log(1-1/n^2) can be proven using the ratio test, as long as the limit of the ratio of consecutive terms is less than one. This confirms that the terms of the series are decreasing and approaching zero, indicating convergence.

5. What are some real-world applications of the convergence of series log(1-1/n^2)?

The convergence of series log(1-1/n^2) has applications in various fields such as statistics, physics, and economics. For example, it can be used to calculate the probability of events in statistics, estimate the behavior of physical systems, and evaluate the growth rate of investments in economics.

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