Finding the Sum of Lns.

Calculus Level 3

What is the value of the following expression:

$\ln\left(1-\frac{1}{2^2}\right)+\ln\left(1-\frac{1}{3^2}\right)+\ln\left(1-\frac{1}{4^2}\right)+\cdots=\, ?$

-\ln (2)

\ln (3)

-2\ln (2)

2\ln (2)

\ln (2)

-\ln (3)

1 solution

X X
Jul 11, 2018

$\ln(1-\dfrac{1}{2^2})+\ln(1-\dfrac{1}{3^2})+\ln(1-\dfrac{1}{4^2})+\dots$

$=\ln\left( \dfrac{1\times3}{2\times2}\times\dfrac{2\times4}{3\times3}\times\dfrac{3\times5}{4\times4}\times\cdots\right)$

$=\ln(\dfrac12)=-\ln2$