Sum of series
- Tolaso J Kos
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Sum of series
Sum the series: $$\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{m^2 n}{3^m \left ( n\cdot 3^m+m\cdot 3^n \right )}$$
Imagination is much more important than knowledge.
Re: Sum of series
Let $$S = \sum _{ m=1 }^{ \infty }{ \sum _{ n=1 }^{ \infty }{ \frac { { m }^{ 2 }n }{ { 3 }^{ m }(n{ 3 }^{ m }+m{ 3 }^{ n }) } } } $$ Using the fact that $$\sum_{m = 1}^\infty\sum_{n = 1}^\infty f(m,n) = \sum_{m = 1}^\infty\sum_{n = 1}^\infty f(n,m) $$ we can rewrite \(S\) as $$S = \sum _{ m=1 }^{ \infty }{ \sum _{ n=1 }^{ \infty }{ \frac { { n }^{ 2 }m }{ { 3 }^{ n }(n{ 3 }^{ m }+m{ 3 }^{ n }) } } } $$ Summing up, we have $$\begin{align*}2S &= \sum_{m = 1}^\infty\sum_{n = 1}^\infty \left({{ \frac { { m }^{ 2 }n }{ { 3 }^{ m }(n{ 3 }^{ m }+m{ 3 }^{ n }) } } } + { \frac { { n }^{ 2 }m }{ { 3 }^{ n }(n{ 3 }^{ m }+m{ 3 }^{ n }) } }\right) \\
&= \sum_{m = 1}^\infty\sum_{n = 1}^\infty\frac{3^nm^2n + 3^mn^2m}{3^{m+n}(n3^m + m3^n)}\\
&= \sum_{m = 1}^\infty\sum_{n = 1}^\infty\frac{mn(m3^n + n3^m)}{3^{m+n}(n3^m + m3^n)}\\
&= \sum_{m = 1}^\infty\sum_{n = 1}^\infty\frac{mn}{3^{m + n}}\\
&= \sum_{m = 1}^\infty\sum_{n = 1}^\infty \left(\frac{m}{3^m}\cdot\frac{n}{3^n}\right)\end{align*}$$ But \(\sum_{m = 1}^\infty\sum_{n = 1}^\infty x_my_n = \left(\sum_{m = 1}^\infty x_m \right)\left(\sum_{n = 1}^\infty y_n \right)\). Hence, $$\begin{align*}2S &= \left(\sum_{m = 1}^\infty \frac{m}{3^m}\right)\left(\sum_{n = 1}^\infty \frac{n}{3^n}\right)\\
&= \left(\sum_{n = 1}^\infty \frac{n}{3^n}\right)^2\end{align*}$$ To find this sum, we apply a classical trick: notice that for \(|x| < 1\),
$$\frac{1}{1-x} = 1 + x + x^2 + \dots$$ Differentiating and multiplying both sides by \(x\), $$\frac{x}{(1-x)^2} = x + 2x^2 + 3x^3 + \dots = \sum_{n = 1}^\infty nx^n$$ Substituting \(x = \frac{1}{3}\), $$\frac{3}{4} = \sum_{n = 1}^\infty \frac{n}{3^n}$$ So we deduce that $$2S = \left(\frac{3}{4}\right)^2$$ and finally, $$\sum _{ m=1 }^{ \infty }{ \sum _{ n=1 }^{ \infty }{ \frac { { m }^{ 2 }n }{ { 3 }^{ m }(n{ 3 }^{ m }+m{ 3 }^{ n }) } } } = S = \frac{9}{32}$$
&= \sum_{m = 1}^\infty\sum_{n = 1}^\infty\frac{3^nm^2n + 3^mn^2m}{3^{m+n}(n3^m + m3^n)}\\
&= \sum_{m = 1}^\infty\sum_{n = 1}^\infty\frac{mn(m3^n + n3^m)}{3^{m+n}(n3^m + m3^n)}\\
&= \sum_{m = 1}^\infty\sum_{n = 1}^\infty\frac{mn}{3^{m + n}}\\
&= \sum_{m = 1}^\infty\sum_{n = 1}^\infty \left(\frac{m}{3^m}\cdot\frac{n}{3^n}\right)\end{align*}$$ But \(\sum_{m = 1}^\infty\sum_{n = 1}^\infty x_my_n = \left(\sum_{m = 1}^\infty x_m \right)\left(\sum_{n = 1}^\infty y_n \right)\). Hence, $$\begin{align*}2S &= \left(\sum_{m = 1}^\infty \frac{m}{3^m}\right)\left(\sum_{n = 1}^\infty \frac{n}{3^n}\right)\\
&= \left(\sum_{n = 1}^\infty \frac{n}{3^n}\right)^2\end{align*}$$ To find this sum, we apply a classical trick: notice that for \(|x| < 1\),
$$\frac{1}{1-x} = 1 + x + x^2 + \dots$$ Differentiating and multiplying both sides by \(x\), $$\frac{x}{(1-x)^2} = x + 2x^2 + 3x^3 + \dots = \sum_{n = 1}^\infty nx^n$$ Substituting \(x = \frac{1}{3}\), $$\frac{3}{4} = \sum_{n = 1}^\infty \frac{n}{3^n}$$ So we deduce that $$2S = \left(\frac{3}{4}\right)^2$$ and finally, $$\sum _{ m=1 }^{ \infty }{ \sum _{ n=1 }^{ \infty }{ \frac { { m }^{ 2 }n }{ { 3 }^{ m }(n{ 3 }^{ m }+m{ 3 }^{ n }) } } } = S = \frac{9}{32}$$
- Tolaso J Kos
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- Posts: 867
- Joined: Sat Nov 07, 2015 6:12 pm
- Location: Larisa
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Re: Sum of series
Jacks , I like your solution very much! Well done !!
Imagination is much more important than knowledge.
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