Matematické Fórum

stuart clark · 08. 01. 2019 14:27

Finding value of $\displaystyle \frac{15}{16}+\frac{15}{16}\cdot\frac{21}{24}+\frac{15}{16}\cdot\frac{21}{24}\cdot \frac{27}{32}+\cdots\cdots$

laszky · 08. 01. 2019 20:17 — Editoval laszky (06. 12. 2023 22:03)

↑ stuart clark:

Hi,

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stuart clark · 21. 01. 2019 14:44

Thanks ↑ laszky:

stuart clark

Evaluation of $\sum^{\infty}_{i=0}\sum^{\infty}_{j=0}\sum^{\infty}_{k=0}\frac{1}{3^i3^j3^k}$ for $i \neq j \neq k$

krakonoš · 27. 01. 2019 21:13

↑ stuart clark:
Hi Stuart Clark
$\sum_{i=0}^{\infty }\sum_{j=0}^{\infty }\sum_{k=0}^{\infty }\frac{1}{3^{i}\cdot3^{j}\cdot 3^{k} }=(\sum_{i=0}^{\infty }\frac{1}{3^{i}})^{3}=\frac{27}{8}$
For i=j=k is
$\sum_{i=0}^{\infty }\sum_{j=0}^{\infty }\sum_{k=0}^{\infty }\frac{1}{3^{i}\cdot 3^{j}\cdot 3^{k}}=\sum_{i=0}^{\infty }(\frac{1}{27})^{i}=\frac{27}{26}$
For $i=j\not =k$ is
$\sum_{i=0}^{\infty }\sum_{j=0}^{\infty }\sum_{k=0}^{\infty }\frac{1}{3^{i}\cdot 3^{j}\cdot 3^{k}}=\sum_{i=0}^{\infty }(\frac{1}{9})^{i}\cdot \sum_{j=0}^{\infty }(\frac{1}{3})^{j}-\sum_{k=0}^{\infty }(\frac{1}{27})^{k}=\frac{27}{16}-\frac{27}{26}$
For $i\not =j\not =k$ is
$\sum_{i=0}^{\infty }\sum_{j=0}^{\infty }\sum_{k=0}^{\infty }\frac{1}{3^{i}\cdot 3^{j}\cdot 3^{k}}=\frac{27}{8}-\frac{27}{26}-3\cdot (\frac{27}{16}-\frac{27}{26})=\frac{81}{208}$