## imqwerty one year ago .......question :)

1. imqwerty

$\sum_{n=1}^{\infty}\left( \frac{ \sqrt{3} }{ 2n } \right)^2 = \left[ \sum_{n=1}^{\infty}\left( \frac{ 1 }{ mn+l} \right) \right]^2$ m,l are integers :) Find m and l

2. inkyvoyd

sir are you sure that the numerator of your second expression isn't (-1)^n? I iz confuzzeld

3. imqwerty

yes i am sure :)

4. anonymous

will check later

5. anonymous

$$\Large \sum_{n=1}^{\infty}\left( \frac{ \sqrt{3} }{ 2n } \right)^2 = 3/4 \sum_{n=1}^{\infty} \frac{ {1} }{ n^2 } =\frac{ {3} }{ 4}.\frac{ {\pi^2} }{ 6}=\frac{ {\pi^2} }{ 2}$$

6. anonymous

$$\Large \left[ \sum_{n=1}^{\infty}\left( \frac{ 1 }{ mn+l} \right) \right]^2 =\dfrac{\pi^2}{2} \\\Large \sum_{n=1}^{\infty}\left( \frac{ 1 }{ mn+l} \right) =\dfrac{\pi^4}{4}$$

7. anonymous

confused as the first series sounds harmonic series @ganeshie8

8. imqwerty

yes..and we don have any summation formula for harmonic series :( there must be some trick

9. anonymous

ok let me think

10. imqwerty

nd that shuld be pi/sqrt{2}

11. anonymous

why so ?

12. imqwerty
13. anonymous

lol yes sorry :3

14. ganeshie8

Clearly there is something wrong with the question. The harmonic series doesn't converge

15. mathmate

Yeah, the first one is $$\pi^2/8$$, and the second one is supposed to diverge whatever finite value we put on l and m. I suspect (as @inkyvoyd did) it's an altenating series, in which case we can get the Leibniz's series when m=1 and l=0 to get $$\pi/4$$. Then a few adjustments should give $$\pi^2/8$$.

16. anonymous

:P yeah for sure

17. anonymous

can we at least share songs lol xD

18. anonymous
19. imqwerty

hey wait there was a typo

20. anonymous

lol ok what the typo is ?

21. mathmate

$$\sum_{n=1}^{\infty}\left( \frac{ \sqrt{3} }{ 2n } \right)^2 = \left[ \sum_{n=1}^{\infty}\left( \frac{ (-1)^n }{ mn+l} \right) \right]^2$$ right?

22. imqwerty

it was meant to be this- $\sum_{n=1}^{\infty}\left( \frac{ \sqrt{3} }{ 2n } \right)=\sum_{n=1}^{\infty}\left( \frac{ 1 }{ mn+l } \right)^2$

23. anonymous

-.- same problem see sqrt3/2n is also harmonic

24. imqwerty

but i know the solution now :D

25. anonymous

really ?

26. anonymous

share

27. imqwerty

ok jst a min :)

28. imqwerty

on the LHS u have -$\sum_{n=1}^{\infty}\frac{ 3 }{ 4n}$take 3/4 out and write it as 1-1/4 nd u get -$\left( 1-\frac{ 1 }{ 4} \right)\left( \sum_{n=1}^{\infty}\frac{ 1 }{ n^2 }\right)$open the brackets nd get this-$\sum_{n=1}^{\infty }\frac{ 1 }{ n^2 } - \sum_{n=1}^{\infty}\frac{ 1 }{ (2n)^2 }$opening the terms u get this kind of series - $1-\frac{ 1 }{ 2^2 }+\frac{ 1 }{ 2^2 } -\frac{ 1 }{ 4^2 }+\frac{ 1 }{ 4^2 }........$+ u get the odd denominator terms too..so finally after cancellations u get this on the LHS- $\sum_{n=1}^{\infty}\frac{ 1 }{ (2n-1)^2 }$ which u can write as- $\sum_{n=1}^{\infty}\left( \frac{ 1 }{ 2n-1 } \right)^2$ so comparing it to the RHS u get m=2 and l= -1

29. anonymous

:)

30. imqwerty

hehe (:

31. mathmate

From what I can see from the solution, the left hand side was meant to be: $$(\sum_{n=1}^{\infty}\left( \frac{ \sqrt{3} }{ 2n } \right))^2=\sum_{n=1}^{\infty}\left( \frac{ 1 }{ mn+l } \right)^2$$

32. imqwerty

yes i jst opened up that square nd wrote the thing initially :)

33. imqwerty

yes @mathmate that shuld be squared sry for so many typos

34. mathmate

That's ok. Too bad LaTex didn't make it much easier!