Express in closed form

**cowboys111** · April 14th 2008, 05:59 PM

Im trying to find a closed form for the nth partial sum of the series ln(1/2)+ln(2/3)+ln(3/4)+...+ln(k/(k+1))+....
My book only has a paragraph on this and i cant understand what they're doing, can someone please try explaing this to me a little better?
Thanks

**PaulRS** · April 14th 2008, 06:05 PM

Ok you want to compute: $\sum_{k=1}^n{\ln\left(\frac{k}{k+1}\right)}$

Note that $\ln\left(\frac{k}{k+1}\right)=\ln(k)-\ln(k+1)$

So: $\sum_{k=1}^n{\left(\ln(k)-\ln(k+1)\right)}$ $=[\ln(1)-\ln(2)]+[\ln(2)-\ln(3)]...+[\ln(n-1)-\ln(n)]+[\ln(n)-\ln(n+1)]$

It's a telescoping sum

, thus shaking off we have: $\sum_{k=1}^n{\left(\ln(k)-\ln(k+1)\right)}=-\ln(n+1)$

**cowboys111** · April 14th 2008, 06:08 PM

That was fast! Thanks for the help

**Soroban** · April 14th 2008, 06:59 PM

Hello, cowboys111!

A slightly different approach . . .

Find a closed form for the $n^{th}$ partial sum of the series:
. . $\ln\left(\frac{1}{2}\right)+\ln\left(\frac{2}{3}\r ight)+\ln\left(\frac{3}{4}\right) + \ln\left(\frac{4}{5}\right) + \hdots$

We have: . $S \;=\;\ln\left(\frac{1}{2}\right) + \ln\left(\frac{2}{3}\right) + \ln\left(\frac{3}{4}\right) + \hdots + \ln\left(\frac{n}{n+1}\right)$

. . $S \;=\;\ln\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\fra c{3}{4}\cdots\frac{n}{n+1}\right) \;=\;\ln\left(\frac{1}{n+1}\right) \;=\; \ln(n+1)^{-1}$

Therefore: . $S \;=\;-\ln(n+1)$

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