imqwerty
  • imqwerty
.......question :)
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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imqwerty
  • 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
inkyvoyd
  • inkyvoyd
sir are you sure that the numerator of your second expression isn't (-1)^n? I iz confuzzeld
imqwerty
  • imqwerty
yes i am sure :)

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anonymous
  • anonymous
will check later
anonymous
  • anonymous
\(\Large \sum_{n=1}^{\infty}\left( \frac{ \sqrt{3} }{ 2n } \right)^2 = 3/4 \sum_{n=1}^{\infty} \frac{ {1} }{ n^2 } =\frac{ {3} }{ 4}.\frac{ {\pi^2} }{ 6}=\frac{ {\pi^2} }{ 2}\)
anonymous
  • anonymous
\(\Large \left[ \sum_{n=1}^{\infty}\left( \frac{ 1 }{ mn+l} \right) \right]^2 =\dfrac{\pi^2}{2} \\\Large \sum_{n=1}^{\infty}\left( \frac{ 1 }{ mn+l} \right) =\dfrac{\pi^4}{4} \)
anonymous
  • anonymous
confused as the first series sounds harmonic series @ganeshie8
imqwerty
  • imqwerty
yes..and we don have any summation formula for harmonic series :( there must be some trick
anonymous
  • anonymous
ok let me think
imqwerty
  • imqwerty
nd that shuld be pi/sqrt{2}
anonymous
  • anonymous
why so ?
imqwerty
  • imqwerty
http://prntscr.com/8ftthq
anonymous
  • anonymous
lol yes sorry :3
ganeshie8
  • ganeshie8
Clearly there is something wrong with the question. The harmonic series doesn't converge
mathmate
  • 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 \(\pi^2/8\).
anonymous
  • anonymous
:P yeah for sure
anonymous
  • anonymous
can we at least share songs lol xD
anonymous
  • anonymous
https://www.youtube.com/watch?v=mWRsgZuwf_8&list=PLPN1tL8FFbmNz7sWLYNyc_D8YFjiVNU3c&index=56
imqwerty
  • imqwerty
hey wait there was a typo
anonymous
  • anonymous
lol ok what the typo is ?
mathmate
  • 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?
imqwerty
  • imqwerty
it was meant to be this- \[\sum_{n=1}^{\infty}\left( \frac{ \sqrt{3} }{ 2n } \right)=\sum_{n=1}^{\infty}\left( \frac{ 1 }{ mn+l } \right)^2\]
anonymous
  • anonymous
-.- same problem see sqrt3/2n is also harmonic
imqwerty
  • imqwerty
but i know the solution now :D
anonymous
  • anonymous
really ?
anonymous
  • anonymous
share
imqwerty
  • imqwerty
ok jst a min :)
imqwerty
  • 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
anonymous
  • anonymous
:)
imqwerty
  • imqwerty
hehe (:
mathmate
  • mathmate
From what I can see from the solution, the left hand side was meant to be: \((\sum_{n=1}^{\infty}\left( \frac{ \sqrt{3} }{ 2n } \right))^2=\sum_{n=1}^{\infty}\left( \frac{ 1 }{ mn+l } \right)^2\)
imqwerty
  • imqwerty
yes i jst opened up that square nd wrote the thing initially :)
imqwerty
  • imqwerty
yes @mathmate that shuld be squared sry for so many typos
mathmate
  • mathmate
That's ok. Too bad LaTex didn't make it much easier!

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