Finding the sum of an infinite series involving logs

[dragoon7201]

∑ln(1-1/(n+1)²)

where n goes from 1 to infinity
I broke the log into 3 parts
ln(n+2) + ln(n) - 2ln(n+1)

but then i don't know how to find the sum of the series.
Can someone enlighten me please

 

[ksdhart]

Just to be sure we're on the same page, the problem you were given is as follows, correct?

\(\displaystyle \sum _{n=1}^{\infty }\:ln\left(1-\frac{1}{\left(x-1\right)^2}\right)\)

 

[dragoon7201]

Yes, that is right, except it is (n+1) not minus 1 . Sorry I couldn't get word to paste properly

 

[daon2]

Notice that all terms are negative and increasing with 0 as the limit of the a_n's, so the integral test will work. It needs to be done by parts and this might be unnecessary for you.

Otherwise, if you know it, the inequality ln(1+t) <= t may be of benefit. Or you can show that ln(1-1/(n+1)^2)/(-1/(n+1)^2) -> 1

 

[MathildaApp]

I think you're on the right track with ln(n+2) + ln(n) - 2ln(n+1) altho not sure how you got there from n(1-1/(n+1)²). Regardless,

sum of ln(n+2) + ln(n) - 2ln(n+1) form n=1 to n= infinity

is the same as:

sum of ln(n) from n=3 to n= infinity
+ sum of ln(n) from n=1 to n= infinity
- sum of ln(n) from n=2 to n= infinity
- sum of ln(n) from n=2 to n= infinity

which is the same as:

sum of ln(n) from n=3 to n= infinity
+ ln(1) + ln(2) + sum of ln(n) from n=3 to n= infinity
- ln(2) - sum of ln(n) from n=3 to n= infinity
- ln(2) - sum of ln(n) from n=3 to n= infinity

and a lot of stuff cancels leaving us with:

ln(1) - ln(2)

 

[soroban]

Hello, dragoon7201!

I agree with MathildaApp.

\(\displaystyle \displaystyle \text{Evaluate: }\:S \;=\; \sum^{\infty}_{n=1}\ln\left[1 - \frac{1}{(n+1)^2}\right] \)

Note that: \(\displaystyle \;\displaystyle 1 - \frac{1}{(n+1)^2} \;=\;\frac{(n+1)^2-1}{(n+1)^2} \;=\; \frac{n^2+2n }{(n+1)^2}\;=\; \frac{n(n+2)}{(n+1)^2}\)

Hence: \(\displaystyle \;\displaystyle S \;=\;\sum^{\infty}_{n=1}\ln\left[\frac{n(n+2)}{(n+1)^2}\right] \)

. . . \(\displaystyle \displaystyle S \;=\;\ln\frac{1\cdot3}{2^2} + \ln\frac{2\cdot4}{3^2} + \ln\frac{3\cdot5}{4^2} + \ln\frac{4\cdot6}{5^2} + \cdots \)

. . .\(\displaystyle \displaystyle S \;=\; \ln\left(\frac{1\cdot3}{2^2}\cdot\frac{2\cdot4}{3^2} \cdot \frac{3\cdot5}{4^2}\cdot\frac{4\cdot6}{5^2} + \cdots \right) \)


We find that everything cancels except the initial \(\displaystyle \frac{1}{2}\)

Therefore: \(\displaystyle \;S \:=\:\ln\left(\frac{1}{2}\right)\)

 

[dragoon7201]

Hello, dragoon7201!

I agree with MathildaApp.


Note that: \(\displaystyle \;\displaystyle 1 - \frac{1}{(n+1)^2} \;=\;\frac{(n+1)^2-1}{(n+1)^2} \;=\; \frac{n^2+2n }{(n+1)^2}\;=\; \frac{n(n+2)}{(n+1)^2}\)

Hence: \(\displaystyle \;\displaystyle S \;=\;\sum^{\infty}_{n=1}\ln\left[\frac{n(n+2)}{(n+1)^2}\right] \)

. . . \(\displaystyle \displaystyle S \;=\;\ln\frac{1\cdot3}{2^2} + \ln\frac{2\cdot4}{3^2} + \ln\frac{3\cdot5}{4^2} + \ln\frac{4\cdot6}{5^2} + \cdots \)

. . .\(\displaystyle \displaystyle S \;=\; \ln\left(\frac{1\cdot3}{2^2}\cdot\frac{2\cdot4}{3^2} \cdot \frac{3\cdot5}{4^2}\cdot\frac{4\cdot6}{5^2} + \cdots \right) \)


We find that everything cancels except the initial \(\displaystyle \frac{1}{2}\)

Therefore: \(\displaystyle \;S \:=\:\ln\left(\frac{1}{2}\right)\)

AHH so this is like a telescopic series. Thanks a lot!