Question

Series question !

Answer 1

Study the nature and find the sum of the following series:
\[\sum_{n=2}^{\infty}\ln(1-\frac{ 1 }{ n^2 })\]

Answer 2

\[S _{n}=\ln(\frac{ 3*8*15*...*n^2-1 }{ 4*9*16*...*n^2 })\]

Answer 3

@ganeshie8 @Hero @Abhisar @.Sam. @Compassionate @nincompoop @eliassaab @ikram002p @Joel_the_boss @Coolsector @One098

Answer 4

http://www.ams.org/bookstore/pspdf/gsm-97-prev.pdf

Answer 5

\[\sum_{n=2}^{\infty}\ln\left(1-\frac{ 1 }{ n^2 }\right) = \ln \prod_{n=2}^{\infty}\left(1-\frac{ 1 }{ n^2 }\right) \]

Answer 6

what's that pi sign never saw that
It's the product ??????

Answer 7

its like sum sign

Answer 8

\[\prod_{n=1}^3 n = 1\times 2 \times 3\]

Answer 9

\[\prod_{n=1}^{10} n = 1\times 2 \times 3\times \cdots \times 10\]

Answer 10

so it's the equivalent of the sum sign for products

Answer 11

Exactly!

Answer 12

but how do you know that relationship ?

Answer 13

just expand it and see

Answer 14

o I saw it actually look at my first posts

Answer 15

we use below log property :
\[ \ln(a) + \ln(b) + \ln(c) + \cdots = \ln(a\times b\times c\times \cdots )\]

Answer 16

sum of logs = log of products

Answer 17

\[\sum_{n=2}^{\infty}\ln\left(1-\frac{ 1 }{ n^2 }\right) = \ln \prod_{n=2}^{\infty}\left(1-\frac{ 1 }{ n^2 }\right) \]

Answer 18

I know I know but i need this: