$\sum_{1\leq l \lt m\lt n} \dfrac{1}{5^l3^m2^n}$

Question

I need a help to deal with the serie $\sum_{1\leq l \lt m\lt n} \dfrac{1}{5^l3^m2^n}$ (OBM level university, 2018). Some ideia? Thanks very much.

Answer 1

\begin{eqnarray*} \sum_{1\le l< m<n}^{K}{\frac{1}{5^l 3^m 2^n}} &=& \sum_{l=1}^{K} {\frac{1}{5^{l}} \sum_{m=l+1}^{K} {\frac{1}{3^{m} }\sum_{n=m+1}^{K} {\frac{1}{2^{n}}} } } \\ &=& \sum_{l=1}^{K} {\frac{1}{5^{l}} \sum_{m=l+1}^{K} {\frac{1}{3^{m}} {\frac{2^{K}-2^{m}}{2^{m+K}} } } } \\ &=& \sum_{l=1}^{K} {\frac{1}{5^{l}} \frac{1}{5} \left( \frac{3^{1-K}}{2^{1+K}}-\frac{3+2}{3^{l} 2^{1+K}} +\frac{1}{6^l} \right) }\\ &=& \frac{1}{203} \frac{1}{2^{3+K} 3^{K} 5^{1+K}} \left(7 \times 2^{3+K} (5\times 3)^{K} -58 \times 3^{K} 5^{k+1} +609\times 5^{K}-375\right) \end{eqnarray*}

For $K\to \infty$, this becomes $\frac{1}{145}$.

Answer 2

Hint: $$\sum_{1\le l<m<n}=\sum_{l=1}^\infty\sum_{m=l+1}^\infty\sum_{n=m+1}^\infty$$ Each sum is a geometric series.

Answer 3

More generally, if $0 < a, b ,c < 1$,

$\begin{array}\\ s(a, b, c) &=\sum_{1\leq l \lt m\lt n} a^lb^mc^n\\ &=\sum_{n=1}^{\infty}\sum_{m=1}^{n-1}\sum_{l=1}^{m-1}a^lb^mc^n\\ &=\sum_{n=1}^{\infty}\sum_{m=1}^{n-1}b^mc^n\sum_{l=1}^{m-1}a^l\\ &=\sum_{n=1}^{\infty}c^n\sum_{m=1}^{n-1}b^m\dfrac{a-a^m}{1-a}\\ &=\dfrac1{1-a}\sum_{n=1}^{\infty}c^n\sum_{m=1}^{n-1}b^m(a-a^m)\\ &=\dfrac1{1-a}\sum_{n=1}^{\infty}c^n\left(a\sum_{m=1}^{n-1}b^m-\sum_{m=1}^{n-1}(ab)^m\right)\\ &=\dfrac1{1-a}\sum_{n=1}^{\infty}c^n\left(a\dfrac{b-b^{n}}{1-b}-\dfrac{(ab)-(ab)^{n}}{1-ab}\right)\\ &=\left(\dfrac1{1-a}\sum_{n=1}^{\infty}c^na\dfrac{b-b^{n}}{1-b}-\dfrac1{1-a}\sum_{n=1}^{\infty}c^n\dfrac{(ab)-(ab)^{n}}{1-ab}\right)\\ &=\dfrac1{(1-a)(1-b)}\sum_{n=1}^{\infty}c^na(b-b^{n})\\ &\quad -\dfrac1{(1-a)(1-ab)}\sum_{n=1}^{\infty}c^n((ab)-(ab)^{n})\\ &=\dfrac1{(1-a)(1-b)}\left(\sum_{n=1}^{\infty}c^nab-\sum_{n=1}^{\infty}c^na(b)^{n}\right)\\ &\quad -\dfrac1{(1-a)(1-ab)}\left(\sum_{n=1}^{\infty}c^nab-\sum_{n=1}^{\infty}c^n(ab)^{n}\right)\\ &=\dfrac1{(1-a)(1-b)}\left(ab\dfrac{c}{1-c}-\dfrac{abc}{1-bc}\right)\\ &\quad -\dfrac1{(1-a)(1-ab)}\left(ab\dfrac{c}{1-c}-\dfrac{abc}{1-abc}\right)\\ &=\dfrac{abc}{(1-a)(1-b)}\left(\dfrac{1}{1-c}-\dfrac{1}{1-bc}\right)\\ &\quad -\dfrac{abc}{(1-a)(1-ab)}\left(\dfrac{1}{1-c}-\dfrac{1}{1-abc}\right)\\ &=\dfrac{abc}{(1-a)(1-b)}\left(\dfrac{c-bc}{(1-c)(1-bc)}\right)\\ &\quad -\dfrac{abc}{(1-a)(1-ab)}\left(\dfrac{c-abc}{(1-c)(1-abc)}\right)\\ &=\dfrac{abc}{(1-a)(1-b)}\left(\dfrac{c(1-b)}{(1-c)(1-bc)}\right)\\ &\quad -\dfrac{abc}{(1-a)(1-ab)}\left(\dfrac{c(1-ab)}{(1-c)(1-abc)}\right)\\ &=\dfrac{abc^2}{(1-a)(1-c)(1-bc)}\\ &\quad -\dfrac{abc^2}{(1-a)(1-c)(1-abc)}\\ &=\dfrac{abc^2}{(1-a)(1-c)(1-bc)((1-abc)}((1-abc)-(1-bc))\\ &=\dfrac{abc^2}{(1-a)(1-c)(1-bc)((1-abc)}(bc-abc)\\ &=\dfrac{abc^2}{(1-a)(1-c)(1-bc)((1-abc)}bc(1-a)\\ &=\dfrac{ab^2c^3}{(1-c)(1-bc)(1-abc)}\\ \end{array} $

If $a=1/5, b=1/3, c=1/2$, this is $\dfrac{1/(5\ 3^2 2^3)}{(1/2)(5/6)(29/30)} =\dfrac{1}{5\ 3^2 2^3(1/2)(5/6)(29/30)} =\dfrac{1}{ 5\ 29} =\dfrac{1}{145} $ which agrees with NivPal's answer.