# How do you find the nth partial sum, determine whether the series converges and find the sum when it exists given ln(1/2)+ln(2/3)+ln(3/4)+...+ln(n/(n+1))+...?

Jan 4, 2018

We know that $\ln \left(\frac{a}{b}\right) = \ln \left(a\right) - \ln \left(b\right)$.

So, we can rewrite the sum
$\ln \left(\frac{1}{2}\right) + \ln \left(\frac{2}{3}\right) + \ln \left(\frac{3}{4}\right) + \ldots + \ln \left(\frac{n - 1}{n}\right) + \ln \left(\frac{n}{n + 1}\right)$
to
=(ln(1)-ln(2))+(ln(2)-ln(3))+(ln(3)-ln(4))+…+(ln(n-1)-ln(n))+(ln(n)-ln(n+1)).

Slightly rearrange the terms:
=ln(1)+(ln(2)-ln(2))+(ln(3)-ln(3))+…+(ln(n-1)-ln(n-1))-ln(n+1)

See how almost all of the terms cancel out? This is called a telescoping sum. Everything simplifies to
$= \ln \left(1\right) - \ln \left(n + 1\right)$
$= \ln \left(\frac{1}{n + 1}\right)$

Now, to find the sum of the infinite series, set $n \to \infty$:
${\lim}_{n \to \infty} \ln \left(\frac{1}{n + 1}\right)$
$= - \infty$

So, the sum of the infinite series diverges.