OpenStudy (imqwerty):

.......question :)

2 years ago
OpenStudy (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 years ago
OpenStudy (inkyvoyd):

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

2 years ago
OpenStudy (imqwerty):

yes i am sure :)

2 years ago
OpenStudy (anonymous):

will check later

2 years ago
OpenStudy (anonymous):

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

2 years ago
OpenStudy (anonymous):

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

2 years ago
OpenStudy (anonymous):

confused as the first series sounds harmonic series @ganeshie8

2 years ago
OpenStudy (imqwerty):

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

2 years ago
OpenStudy (anonymous):

ok let me think

2 years ago
OpenStudy (imqwerty):

nd that shuld be pi/sqrt{2}

2 years ago
OpenStudy (anonymous):

why so ?

2 years ago
OpenStudy (imqwerty):

http://prntscr.com/8ftthq

2 years ago
OpenStudy (anonymous):

lol yes sorry :3

2 years ago
ganeshie8 (ganeshie8):

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

2 years ago
OpenStudy (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 \(\pi^2/8\).

2 years ago
OpenStudy (anonymous):

:P yeah for sure

2 years ago
OpenStudy (anonymous):

can we at least share songs lol xD

2 years ago
OpenStudy (imqwerty):

hey wait there was a typo

2 years ago
OpenStudy (anonymous):

lol ok what the typo is ?

2 years ago
OpenStudy (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?

2 years ago
OpenStudy (imqwerty):

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

2 years ago
OpenStudy (anonymous):

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

2 years ago
OpenStudy (imqwerty):

but i know the solution now :D

2 years ago
OpenStudy (anonymous):

really ?

2 years ago
OpenStudy (anonymous):

share

2 years ago
OpenStudy (imqwerty):

ok jst a min :)

2 years ago
OpenStudy (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

2 years ago
OpenStudy (anonymous):

:)

2 years ago
OpenStudy (imqwerty):

hehe (:

2 years ago
OpenStudy (mathmate):

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

2 years ago
OpenStudy (imqwerty):

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

2 years ago
OpenStudy (imqwerty):

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

2 years ago
OpenStudy (mathmate):

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

2 years ago
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