I found a straight solution without even symmetry, and I think this solutions makes more sense than the symmetry one. $S$ can be re-written as
\[\sum_{n=1}^\infty\sum_{n=1}^\infty \dfrac{n^2n}{3^n\cdot(n3^n+n3^n)}=\dfrac12\sum_{n=1}^\infty\sum_{n=1}^\infty \dfrac{n^2}{\left(3^n\right)^2}=\dfrac12\sum_{n=1}^\infty\sum_{n=1}^\infty \left(\dfrac{n}{3^n}\right)\left(\dfrac{n}{3^n}\right)=\dfrac12\left(\sum_{n=1}^\infty\dfrac n{3^n}\right)^2\]
Note the trick to write the sum as the last portion. It is actually the following:
\[\sum_{i=1}^k\sum_{j=1}^la_ib_j=\left(\sum_{i=1}^ka_i\right)\left(\sum_{j=1}^lb_i\right)\]
Now $T=\sum\limits_{n=1}^\infty\dfrac n{3^n}$ which can be found easily.
Re: BdMO National Higher Secondary 2009/11
Posted: Thu Jan 23, 2014 8:55 pm
by *Mahi*
Masum wrote:I found a straight solution without even symmetry, and I think this solutions makes more sense than the symmetry one. $S$ can be re-written as
\[\sum_{n=1}^\infty\sum_{n=1}^\infty \dfrac{n^2n}{3^n\cdot(n3^n+n3^n)}=\dfrac12\sum_{n=1}^\infty\sum_{n=1}^\infty \dfrac{n^2}{\left(3^n\right)^2}=\dfrac12\sum_{n=1}^\infty\sum_{n=1}^\infty \left(\dfrac{n}{3^n}\right)\left(\dfrac{n}{3^n}\right)=\dfrac12\left(\sum_{n=1}^\infty\dfrac n{3^n}\right)^2\]
Note the trick to write the sum as the last portion. It is actually the following:
\[\sum_{i=1}^k\sum_{j=1}^la_ib_j=\left(\sum_{i=1}^ka_i\right)\left(\sum_{j=1}^lb_i\right)\]
Now $T=\sum\limits_{n=1}^\infty\dfrac n{3^n}$ which can be found easily.
I am not quite sure about this :/
For example, let $\{a_i \} = \{i\}$ and $\{b_i \} = \{\frac 1 i\}$.
Then according to your method \[\sum _{i=1}^{n} \sum _{j=1}^{n} a_ib_j = \sum _{i=1}^{n} \sum _{i=1}^{n} a_ib_i = \sum _{i=1}^{n} \sum _{i=1}^{n} 1 = \sum _{i=1}^{n} \sum _{i=1}^{n} 1 \cdot 1 = \left (\sum _{i=1}^{n} 1 \right)^2 = n^2 \]
While \[\sum _{i=1}^{n} \sum _{j=1}^{n} a_ib_j = \left( \sum _{i=1}^{n} i \right) \left( \sum _{i=1}^{n} \frac 1 i \right) \] which is definitely not $n^2$
Re: BdMO National Higher Secondary 2009/11
Posted: Fri Jan 24, 2014 9:25 am
by Masum
Hmm. May be a miss-statement. This is infact a generalization of $a(c+d)+b(c+d)=(a+b)(c+d)$ so far as I am concerned.
Re: BdMO National Higher Secondary 2009/11
Posted: Fri Jan 24, 2014 1:13 pm
by *Mahi*
Masum wrote:Hmm. May be a miss-statement. This is infact a generalization of $a(c+d)+b(c+d)=(a+b)(c+d)$ so far as I am concerned.
$\sum _{i=1}^{n} \sum _{j=1}^{n} a_ib_j = (\sum _{i=1}^{n} a_i) (\sum _{j=1}^{n} b_j)$ might be, but I am not sure about $\sum _{i=1}^{n} \sum _{j=1}^{n} a_ib_j = \sum _{i=1}^{n} \sum _{i=1}^{n} a_ib_i $