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Need serious series help again..

Mathematics
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@Mertsj can u help?

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Other answers:

did you plugged in few values ... nice series !
ohh yea.. hold on, tnx
easy to prove by induction
zarkon, more hints? i've solved like 40 series problems today, my brains are malfunctioning.
use induction...prove for n=1...assume true of n...show true for n+1
you can also do a direct derivation...
\[\sum _{k=1}^n \text{ln}\left[\frac{k(k+2)}{(k+1)^2}\right]\] \[=\text{ln}\left[\prod _{k=1}^n \frac{k(k+2)}{(k+1)^2}\right]\] then show \[\prod _{k=1}^n \frac{k(k+2)}{(k+1)^2}=\frac{n+2}{2(n+1)}\]
\[\prod _{k=1}^n \frac{k(k+2)}{(k+1)^2}\] \[=\frac{(1\cdot2\cdot3\cdot4\cdots n)(3\cdot4\cdot5\cdots n\cdot (n+1)\cdot(n+2)}{2^2\cdot3^2\cdot4^2\cdots n^2\cdot(n+1)^2}\]
\[=\frac{(1\cdot2\cdot\cancel{3}\cdot\cancel{4}\cdots \cancel{n})(\cancel{3}\cdot\cancel{4}\cdot\cancel{5}\cdots \cancel{n}\cdot (n+1)\cdot(n+2)}{2^2\cdot\cancel{3^2}\cdot\cancel{4^2}\cdots \cancel{n^2}\cdot(n+1)^2}\] \[=\frac{2(n+1)(n+2)}{2^2(n+1)^2}=\frac{n+2}{2(n+1)}\]
see what I did?
yes, TY!
yw
fun problem

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