# Thread: closed form of nth partial sum

1. ## closed form of nth partial sum

find a closed form for the nth partial sum of the series and determine whether the series converges. If so, find its sum.

ln(1 - (1/4)) + ln(1 - (1/9)) + ln(1 - (1/16)) +...+ ln(1 - (1/(k+1)^2)+...

using log properties i got the first few partial sums as S1 = 3/4, S2 = 2/3, S3 = 5/8, S4 = 3/5 but i can't seem to find a pattern. also in the answers in the back of the book, even though this question asked for a closed form, the back of the book gave: sum from k=2 to (n+1) of [ln((k-1)/(k)) - ln(k/(k+1))]. how did the book get that answer?

2. Originally Posted by oblixps
find a closed form for the nth partial sum of the series and determine whether the series converges. If so, find its sum.

ln(1 - (1/4)) + ln(1 - (1/9)) + ln(1 - (1/16)) +...+ ln(1 - (1/(k+1)^2)+...

using log properties i got the first few partial sums as S1 = 3/4, S2 = 2/3, S3 = 5/8, S4 = 3/5 but i can't seem to find a pattern. also in the answers in the back of the book, even though this question asked for a closed form, the back of the book gave: sum from k=2 to (n+1) of [ln((k-1)/(k)) - ln(k/(k+1))]. i thought closed form meant without a summation symbol. i am confused.
$\sum_{k=2}^{m}\ln\left(1-\frac{1}{k^2}\right)=\sum_{k=2}^{m}\ln\left(\frac{ k^2-1}{k^2}\right)=$ $\sum_{k=2}^{m}\left\{\ln(k^2-1)-\ln(k^2)\right)=-\sum_{k=2}^{m}\left\{\ln(k^2)-\ln(k^2-1)\right\}$

3. i don't understand how you got that. could you explain how your first step came about?

4. Originally Posted by oblixps
i don't understand how you got that. could you explain how your first step came about?
$1-\frac{1}{k^2}=\frac{k^2}{k^2}-\frac{1}{k^2}=\frac{k^2-1}{k^2}$

5. oh wait i understand it now. silly me. thanks!

6. The exact value of...

$S = \lim_{m \rightarrow \infty} \sum_{k=2}^{m} \ln (1-\frac{1}{k^{2}})$ (1)

... can be found as follows. Starting from the well known 'infinite product' ...

$\frac{\sin \pi x}{\pi x}= \prod_{k=1}^{\infty} (1-\frac{x^{2}}{k^{2}})$ (2)

... we obtain first...

$\prod_{k=2}^{\infty} (1-\frac{x^{2}}{k^{2}}) = \frac{\sin \pi x}{\pi x (1-x^{2})}$ (3)

... and then...

$\prod_{k=2}^{\infty} (1-\frac{1}{k^{2}}) = \lim_{x \rightarrow 1} \frac{\sin \pi x}{\pi x (1-x^{2})} = \lim_{x \rightarrow 1} \frac{\cos \pi x}{(1-3 x^{2})} = \frac{1}{2}$ (4)

... so that...

$S = \lim_{m \rightarrow \infty} \sum_{k=2}^{m} \ln (1-\frac{1}{k^{2}})= - \ln 2$ (5)

Kind regards

$\chi$ $\sigma$