Question

.......question :)

Answer 1

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

m,l are integers :)
Find m and l

Answer 2

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

Answer 3

yes i am sure :)

Answer 4

will check later

Answer 5

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

Answer 6

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

Answer 7

confused as the first series sounds harmonic series @ganeshie8

Answer 8

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

Answer 9

ok let me think

Answer 10

nd that shuld be pi/sqrt{2}

Answer 11

why so ?

Answer 12

http://prntscr.com/8ftthq

Answer 13

lol yes sorry :3

Answer 14

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

Answer 15

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 \(\pi^2/8\).

Answer 16

:P yeah for sure

Answer 17

can we at least share songs lol xD

Answer 18

https://www.youtube.com/watch?v=mWRsgZuwf_8&list=PLPN1tL8FFbmNz7sWLYNyc_D8YFjiVNU3c&index=56

Answer 19

hey wait
there was a typo

Answer 20

lol ok what the typo is ?

Answer 21

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

Answer 22

it was meant to be this-

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

Answer 23

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

Answer 24

but i know the solution now :D

Answer 25

really ?

Answer 26

share

Answer 27

ok jst a min :)

Answer 28

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

Answer 29

:)

Answer 30

hehe (:

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

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