bahrom7893
  • bahrom7893
Need serious series help again..
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
bahrom7893
  • bahrom7893
|dw:1330836048396:dw|
bahrom7893
  • bahrom7893
Show that: |dw:1330836111973:dw|
bahrom7893
  • bahrom7893
@Mertsj can u help?

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More answers

anonymous
  • anonymous
did you plugged in few values ... nice series !
bahrom7893
  • bahrom7893
ohh yea.. hold on, tnx
Zarkon
  • Zarkon
easy to prove by induction
bahrom7893
  • bahrom7893
zarkon, more hints? i've solved like 40 series problems today, my brains are malfunctioning.
Zarkon
  • Zarkon
use induction...prove for n=1...assume true of n...show true for n+1
Zarkon
  • Zarkon
you can also do a direct derivation...
Zarkon
  • Zarkon
\[\sum _{k=1}^n \text{ln}\left[\frac{k(k+2)}{(k+1)^2}\right]\] \[=\text{ln}\left[\prod _{k=1}^n \frac{k(k+2)}{(k+1)^2}\right]\] then show \[\prod _{k=1}^n \frac{k(k+2)}{(k+1)^2}=\frac{n+2}{2(n+1)}\]
Zarkon
  • Zarkon
\[\prod _{k=1}^n \frac{k(k+2)}{(k+1)^2}\] \[=\frac{(1\cdot2\cdot3\cdot4\cdots n)(3\cdot4\cdot5\cdots n\cdot (n+1)\cdot(n+2)}{2^2\cdot3^2\cdot4^2\cdots n^2\cdot(n+1)^2}\]
Zarkon
  • Zarkon
\[=\frac{(1\cdot2\cdot\cancel{3}\cdot\cancel{4}\cdots \cancel{n})(\cancel{3}\cdot\cancel{4}\cdot\cancel{5}\cdots \cancel{n}\cdot (n+1)\cdot(n+2)}{2^2\cdot\cancel{3^2}\cdot\cancel{4^2}\cdots \cancel{n^2}\cdot(n+1)^2}\] \[=\frac{2(n+1)(n+2)}{2^2(n+1)^2}=\frac{n+2}{2(n+1)}\]
Zarkon
  • Zarkon
see what I did?
bahrom7893
  • bahrom7893
yes, TY!
Zarkon
  • Zarkon
yw
Zarkon
  • Zarkon
fun problem

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