How to find the sum of another series
How do I find the sum of a series
$$sum_{1 leq x < y < z}^infty frac{1}{3^x4^y5^z}$$
sequences-and-series
asked Dec 24 at 20:49
Some student
62
add a comment |
How do I find the sum of a series
$$sum_{1 leq x < y < z}^infty frac{1}{3^x4^y5^z}$$
sequences-and-series
asked Dec 24 at 20:49
Some student
62
Try something along these lines.
– Clement C.
Dec 24 at 20:52
1
Who DELETED math.stackexchange.com/questions/3050493??
– cgiovanardi
Dec 25 at 15:24
add a comment |
How do I find the sum of a series
$$sum_{1 leq x < y < z}^infty frac{1}{3^x4^y5^z}$$
sequences-and-series
asked Dec 24 at 20:49
Some student
62
How do I find the sum of a series
$$sum_{1 leq x < y < z}^infty frac{1}{3^x4^y5^z}$$
sequences-and-series
sequences-and-series
asked Dec 24 at 20:49
Some student
62
asked Dec 24 at 20:49
Some student
62
asked Dec 24 at 20:49
Some student
62
asked Dec 24 at 20:49
Some student
62
asked Dec 24 at 20:49
Some student
62
62
Try something along these lines.
– Clement C.
Dec 24 at 20:52
1
Who DELETED math.stackexchange.com/questions/3050493??
– cgiovanardi
Dec 25 at 15:24
add a comment |
Try something along these lines.
– Clement C.
Dec 24 at 20:52
1
Who DELETED math.stackexchange.com/questions/3050493??
– cgiovanardi
Dec 25 at 15:24
Try something along these lines.
– Clement C.
Dec 24 at 20:52
Try something along these lines.
– Clement C.
Dec 24 at 20:52
1
1
Who DELETED math.stackexchange.com/questions/3050493??
– cgiovanardi
Dec 25 at 15:24
Who DELETED math.stackexchange.com/questions/3050493??
– cgiovanardi
Dec 25 at 15:24
add a comment |
2 Answers
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Write $i=x,,j=y-x,,k=z-y$ so your sum is $sum_{ijk}frac{1}{60^i 20^j 5^k}$, all indices starting at $1$. But $sum_ifrac{1}{n^i}=frac{1}{n-1}$, so the result is $frac{1}{59times 19times 4}$.
answered Dec 24 at 20:54
J.G.
22.4k22035
add a comment |
Direct calculation through expansion:
$$begin{align}sum_{1 leq x < y < z}^infty frac{1}{3^x4^y5^z}=
&color{red}{frac13left[frac1{4^2}left(frac1{5^3}+frac1{5^4}+cdotsright)+frac1{4^3}left(frac1{5^4}+frac1{5^5}+cdotsright)+cdotsright]}+\
+&color{blue}{frac1{3^2}left[frac1{4^3}left(frac1{5^4}+frac1{5^5}+cdotsright)+frac1{4^4}left(frac1{5^5}+frac1{5^6}+cdotsright)+cdotsright]}+cdots=\
=&color{red}{frac13left[frac1{4^2}cdot frac{1}{4cdot 5^2}+frac1{4^3}cdot frac1{4cdot 5^3}+cdotsright]}+\
+&color{blue}{frac1{3^2}left[frac1{4^3}cdot frac1{4cdot 5^3}+frac1{4^4}cdot frac1{4cdot 5^4}+cdotsright]}+cdots=\
=&color{red}{frac13cdot frac{1}{4^3cdot 5^2}cdot frac{4cdot 5}{19}}+\
+&color{blue}{frac1{3^2}cdot frac{1}{4^4cdot 5^3}cdot frac{4cdot 5}{19}}+cdots=\
=&frac{1}{3cdot 4^2cdot 5cdot 19}cdot frac{3cdot 4cdot 5}{59}=\
=&frac1{4cdot 19cdot 59}.end{align}$$
answered 2 days ago
farruhota
19.1k2736
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
Write $i=x,,j=y-x,,k=z-y$ so your sum is $sum_{ijk}frac{1}{60^i 20^j 5^k}$, all indices starting at $1$. But $sum_ifrac{1}{n^i}=frac{1}{n-1}$, so the result is $frac{1}{59times 19times 4}$.
answered Dec 24 at 20:54
J.G.
22.4k22035
add a comment |
Write $i=x,,j=y-x,,k=z-y$ so your sum is $sum_{ijk}frac{1}{60^i 20^j 5^k}$, all indices starting at $1$. But $sum_ifrac{1}{n^i}=frac{1}{n-1}$, so the result is $frac{1}{59times 19times 4}$.
answered Dec 24 at 20:54
J.G.
22.4k22035
add a comment |
Write $i=x,,j=y-x,,k=z-y$ so your sum is $sum_{ijk}frac{1}{60^i 20^j 5^k}$, all indices starting at $1$. But $sum_ifrac{1}{n^i}=frac{1}{n-1}$, so the result is $frac{1}{59times 19times 4}$.
answered Dec 24 at 20:54
J.G.
22.4k22035
Write $i=x,,j=y-x,,k=z-y$ so your sum is $sum_{ijk}frac{1}{60^i 20^j 5^k}$, all indices starting at $1$. But $sum_ifrac{1}{n^i}=frac{1}{n-1}$, so the result is $frac{1}{59times 19times 4}$.
answered Dec 24 at 20:54
J.G.
22.4k22035
answered Dec 24 at 20:54
J.G.
22.4k22035
answered Dec 24 at 20:54
J.G.
22.4k22035
answered Dec 24 at 20:54
J.G.
22.4k22035
22.4k22035
add a comment |
add a comment |
Direct calculation through expansion:
$$begin{align}sum_{1 leq x < y < z}^infty frac{1}{3^x4^y5^z}=
&color{red}{frac13left[frac1{4^2}left(frac1{5^3}+frac1{5^4}+cdotsright)+frac1{4^3}left(frac1{5^4}+frac1{5^5}+cdotsright)+cdotsright]}+\
+&color{blue}{frac1{3^2}left[frac1{4^3}left(frac1{5^4}+frac1{5^5}+cdotsright)+frac1{4^4}left(frac1{5^5}+frac1{5^6}+cdotsright)+cdotsright]}+cdots=\
=&color{red}{frac13left[frac1{4^2}cdot frac{1}{4cdot 5^2}+frac1{4^3}cdot frac1{4cdot 5^3}+cdotsright]}+\
+&color{blue}{frac1{3^2}left[frac1{4^3}cdot frac1{4cdot 5^3}+frac1{4^4}cdot frac1{4cdot 5^4}+cdotsright]}+cdots=\
=&color{red}{frac13cdot frac{1}{4^3cdot 5^2}cdot frac{4cdot 5}{19}}+\
+&color{blue}{frac1{3^2}cdot frac{1}{4^4cdot 5^3}cdot frac{4cdot 5}{19}}+cdots=\
=&frac{1}{3cdot 4^2cdot 5cdot 19}cdot frac{3cdot 4cdot 5}{59}=\
=&frac1{4cdot 19cdot 59}.end{align}$$
answered 2 days ago
farruhota
19.1k2736
add a comment |
Direct calculation through expansion:
$$begin{align}sum_{1 leq x < y < z}^infty frac{1}{3^x4^y5^z}=
&color{red}{frac13left[frac1{4^2}left(frac1{5^3}+frac1{5^4}+cdotsright)+frac1{4^3}left(frac1{5^4}+frac1{5^5}+cdotsright)+cdotsright]}+\
+&color{blue}{frac1{3^2}left[frac1{4^3}left(frac1{5^4}+frac1{5^5}+cdotsright)+frac1{4^4}left(frac1{5^5}+frac1{5^6}+cdotsright)+cdotsright]}+cdots=\
=&color{red}{frac13left[frac1{4^2}cdot frac{1}{4cdot 5^2}+frac1{4^3}cdot frac1{4cdot 5^3}+cdotsright]}+\
+&color{blue}{frac1{3^2}left[frac1{4^3}cdot frac1{4cdot 5^3}+frac1{4^4}cdot frac1{4cdot 5^4}+cdotsright]}+cdots=\
=&color{red}{frac13cdot frac{1}{4^3cdot 5^2}cdot frac{4cdot 5}{19}}+\
+&color{blue}{frac1{3^2}cdot frac{1}{4^4cdot 5^3}cdot frac{4cdot 5}{19}}+cdots=\
=&frac{1}{3cdot 4^2cdot 5cdot 19}cdot frac{3cdot 4cdot 5}{59}=\
=&frac1{4cdot 19cdot 59}.end{align}$$
answered 2 days ago
farruhota
19.1k2736
add a comment |
Direct calculation through expansion:
$$begin{align}sum_{1 leq x < y < z}^infty frac{1}{3^x4^y5^z}=
&color{red}{frac13left[frac1{4^2}left(frac1{5^3}+frac1{5^4}+cdotsright)+frac1{4^3}left(frac1{5^4}+frac1{5^5}+cdotsright)+cdotsright]}+\
+&color{blue}{frac1{3^2}left[frac1{4^3}left(frac1{5^4}+frac1{5^5}+cdotsright)+frac1{4^4}left(frac1{5^5}+frac1{5^6}+cdotsright)+cdotsright]}+cdots=\
=&color{red}{frac13left[frac1{4^2}cdot frac{1}{4cdot 5^2}+frac1{4^3}cdot frac1{4cdot 5^3}+cdotsright]}+\
+&color{blue}{frac1{3^2}left[frac1{4^3}cdot frac1{4cdot 5^3}+frac1{4^4}cdot frac1{4cdot 5^4}+cdotsright]}+cdots=\
=&color{red}{frac13cdot frac{1}{4^3cdot 5^2}cdot frac{4cdot 5}{19}}+\
+&color{blue}{frac1{3^2}cdot frac{1}{4^4cdot 5^3}cdot frac{4cdot 5}{19}}+cdots=\
=&frac{1}{3cdot 4^2cdot 5cdot 19}cdot frac{3cdot 4cdot 5}{59}=\
=&frac1{4cdot 19cdot 59}.end{align}$$
answered 2 days ago
farruhota
19.1k2736
Direct calculation through expansion:
$$begin{align}sum_{1 leq x < y < z}^infty frac{1}{3^x4^y5^z}=
&color{red}{frac13left[frac1{4^2}left(frac1{5^3}+frac1{5^4}+cdotsright)+frac1{4^3}left(frac1{5^4}+frac1{5^5}+cdotsright)+cdotsright]}+\
+&color{blue}{frac1{3^2}left[frac1{4^3}left(frac1{5^4}+frac1{5^5}+cdotsright)+frac1{4^4}left(frac1{5^5}+frac1{5^6}+cdotsright)+cdotsright]}+cdots=\
=&color{red}{frac13left[frac1{4^2}cdot frac{1}{4cdot 5^2}+frac1{4^3}cdot frac1{4cdot 5^3}+cdotsright]}+\
+&color{blue}{frac1{3^2}left[frac1{4^3}cdot frac1{4cdot 5^3}+frac1{4^4}cdot frac1{4cdot 5^4}+cdotsright]}+cdots=\
=&color{red}{frac13cdot frac{1}{4^3cdot 5^2}cdot frac{4cdot 5}{19}}+\
+&color{blue}{frac1{3^2}cdot frac{1}{4^4cdot 5^3}cdot frac{4cdot 5}{19}}+cdots=\
=&frac{1}{3cdot 4^2cdot 5cdot 19}cdot frac{3cdot 4cdot 5}{59}=\
=&frac1{4cdot 19cdot 59}.end{align}$$
answered 2 days ago
farruhota
19.1k2736
answered 2 days ago
farruhota
19.1k2736
answered 2 days ago
farruhota
19.1k2736
answered 2 days ago
farruhota
19.1k2736
19.1k2736
add a comment |
add a comment |
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Try something along these lines.
– Clement C.
Dec 24 at 20:52
1
Who DELETED math.stackexchange.com/questions/3050493??
– cgiovanardi
Dec 25 at 15:24