# Closed form of a partial sum?

• Apr 1st 2010, 04:15 AM
TripleHelix
Closed form of a partial sum?
Hey fellow math people,

I'm Matt, new to the board and looking forward to reading posts and learning as much as I can so I can get through my calc II class and beyond.

To start, I'm having trouble with finding the closed form of an nth partial sum of an infinite series. Here's the question given:

ln(1/2) + ln(2/3) + ln(3/4) +...+ ln(k/k+1) +...

Directions say to find closed form and if it converges, find limit.

Any help appreciated, I'm having trouble comprehending the process of finding a closed form. I know you subtract (Sn) from some other form of the series--but what's the common ratio betwee the two? Or am I going off on a tangent...

Thanks!
Matt
• Apr 1st 2010, 04:20 AM
Prove It
Quote:

Originally Posted by TripleHelix
Hey fellow math people,

I'm Matt, new to the board and looking forward to reading posts and learning as much as I can so I can get through my calc II class and beyond.

To start, I'm having trouble with finding the closed form of an nth partial sum of an infinite series. Here's the question given:

ln(1/2) + ln(2/3) + ln(3/4) +...+ ln(k/k+1) +...

Directions say to find closed form and if it converges, find limit.

Any help appreciated, I'm having trouble comprehending the process of finding a closed form. I know you subtract (Sn) from some other form of the series--but what's the common ratio betwee the two? Or am I going off on a tangent...

Thanks!
Matt

Notice that

$\ln{\frac{p}{q}} = \ln{p} - \ln{q}$.

So that means:

$\ln{\frac{1}{2}} + \ln{\frac{2}{3}} + \ln{\frac{3}{4}} + \dots + \ln{\frac{k}{k + 1}}$

$= \ln{1} - \ln{2} + \ln{2} - \ln{3} + \ln{3} - \ln{4} + \dots + \ln{k} - \ln{(k + 1)}$

$= \ln{1} - \ln{(k + 1)}$

$= 0 - \ln{(k + 1)}$

$= \ln{\left(\frac{1}{k + 1}\right)}$.

So what do you think happens as $k \to \infty$?
• Apr 1st 2010, 04:43 AM
TripleHelix
I'm gonna say it diverges because it just gets infinitely larger.

Thanks, the natural log property you showed clears it up. I'm an English major and have been out of the math loop for awhile, so I'm lacking on my algebra skills.
• Apr 1st 2010, 06:35 AM
HallsofIvy
A slightly different way: instead of separating the logs more, combine them:

$ln(\frac{1}{2})+ ln(\frac{2}{3})+ ln(\frac{3}{4})+ \cdot\cdot\cdot+ ln(\frac{k}{k+1})= ln(\frac{1}{2}\frac{2}{3}\frac{3}{4}\cdot\cdot\cdo t\frac{k}{k+1})= ln(\frac{1}{k+1})$

Notice that, for all k>1, $\frac{1}{k+1}< 1$ so $ln(\frac{1}{k+1})< 0$.
• Apr 1st 2010, 05:59 PM
Prove It
Thanks HallsofIvy, I've fixed up the typo in mine...
• Apr 1st 2010, 06:52 PM
Krizalid
mm $\ln\left(\frac k{k+1}\right)=\ln k-\ln(k+1),$ telescoping sum, that's all, and the series diverges.

ahhh, it was proved above, well.
• Apr 1st 2010, 06:56 PM
TripleHelix
Thanks everybody.

The answer my book gives is:

Sn= -ln(n+1) and the limit is -infinity.