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

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imqwerty
 one year ago
Best ResponseYou've already chosen the best response.4\[\sum_{n=1}^{\infty}\left( \frac{ \sqrt{3} }{ 2n } \right)^2 = \left[ \sum_{n=1}^{\infty}\left( \frac{ 1 }{ mn+l} \right) \right]^2\] m,l are integers :) Find m and l

inkyvoyd
 one year ago
Best ResponseYou've already chosen the best response.1sir are you sure that the numerator of your second expression isn't (1)^n? I iz confuzzeld

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\Large \sum_{n=1}^{\infty}\left( \frac{ \sqrt{3} }{ 2n } \right)^2 = 3/4 \sum_{n=1}^{\infty} \frac{ {1} }{ n^2 } =\frac{ {3} }{ 4}.\frac{ {\pi^2} }{ 6}=\frac{ {\pi^2} }{ 2}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\Large \left[ \sum_{n=1}^{\infty}\left( \frac{ 1 }{ mn+l} \right) \right]^2 =\dfrac{\pi^2}{2} \\\Large \sum_{n=1}^{\infty}\left( \frac{ 1 }{ mn+l} \right) =\dfrac{\pi^4}{4} \)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0confused as the first series sounds harmonic series @ganeshie8

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.4yes..and we don have any summation formula for harmonic series :( there must be some trick

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.4nd that shuld be pi/sqrt{2}

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Clearly there is something wrong with the question. The harmonic series doesn't converge

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, 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
 one year ago
Best ResponseYou've already chosen the best response.0can we at least share songs lol xD

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0https://www.youtube.com/watch?v=mWRsgZuwf_8&list=PLPN1tL8FFbmNz7sWLYNyc_D8YFjiVNU3c&index=56

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.4hey wait there was a typo

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lol ok what the typo is ?

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0\(\sum_{n=1}^{\infty}\left( \frac{ \sqrt{3} }{ 2n } \right)^2 = \left[ \sum_{n=1}^{\infty}\left( \frac{ (1)^n }{ mn+l} \right) \right]^2\) right?

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.4it 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
 one year ago
Best ResponseYou've already chosen the best response.0. same problem see sqrt3/2n is also harmonic

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.4but i know the solution now :D

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.4on the LHS u have \[\sum_{n=1}^{\infty}\frac{ 3 }{ 4n}\]take 3/4 out and write it as 11/4 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 \[\sum_{n=1}^{\infty}\frac{ 1 }{ (2n1)^2 } \] which u can write as \[\sum_{n=1}^{\infty}\left( \frac{ 1 }{ 2n1 } \right)^2 \] so comparing it to the RHS u get m=2 and l= 1

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0From 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
 one year ago
Best ResponseYou've already chosen the best response.4yes i jst opened up that square nd wrote the thing initially :)

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.4yes @mathmate that shuld be squared sry for so many typos

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0That's ok. Too bad LaTex didn't make it much easier!
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