## bahrom7893 3 years ago Need serious series help again..

1. bahrom7893

|dw:1330836048396:dw|

2. bahrom7893

Show that: |dw:1330836111973:dw|

3. bahrom7893

@Mertsj can u help?

4. mathg8

did you plugged in few values ... nice series !

5. bahrom7893

ohh yea.. hold on, tnx

6. Zarkon

easy to prove by induction

7. bahrom7893

zarkon, more hints? i've solved like 40 series problems today, my brains are malfunctioning.

8. Zarkon

use induction...prove for n=1...assume true of n...show true for n+1

9. Zarkon

you can also do a direct derivation...

10. 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 $\prod _{k=1}^n \frac{k(k+2)}{(k+1)^2}=\frac{n+2}{2(n+1)}$

11. Zarkon

$\prod _{k=1}^n \frac{k(k+2)}{(k+1)^2}$ $=\frac{(1\cdot2\cdot3\cdot4\cdots n)(3\cdot4\cdot5\cdots n\cdot (n+1)\cdot(n+2)}{2^2\cdot3^2\cdot4^2\cdots n^2\cdot(n+1)^2}$

12. Zarkon

$=\frac{(1\cdot2\cdot\cancel{3}\cdot\cancel{4}\cdots \cancel{n})(\cancel{3}\cdot\cancel{4}\cdot\cancel{5}\cdots \cancel{n}\cdot (n+1)\cdot(n+2)}{2^2\cdot\cancel{3^2}\cdot\cancel{4^2}\cdots \cancel{n^2}\cdot(n+1)^2}$ $=\frac{2(n+1)(n+2)}{2^2(n+1)^2}=\frac{n+2}{2(n+1)}$

13. Zarkon

see what I did?

14. bahrom7893

yes, TY!

15. Zarkon

yw

16. Zarkon

fun problem