1
$\begingroup$

We can find the sum of infinite geometric series but I am stuck on this problem. Find the sum of the following infinite series:

$$\frac16+\frac{5}{6\cdot 12} + \frac{5\cdot8}{6\cdot12\cdot18} + \frac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+\dots$$

$\endgroup$
5
$\begingroup$

Using binomial expansion, we have:

$(1-x)^{-2/3} = 1 + \dfrac{\frac{2}{3}}{1!}x + \dfrac{\frac{2}{3} \cdot \frac{5}{3}}{2!} x^2 + \dfrac{\frac{2}{3} \cdot \frac{5}{3} \cdot \frac{8}{3}}{3!}x^3 + \dfrac{\frac{2}{3} \cdot \frac{5}{3} \cdot \frac{8}{3} \cdot \frac{11}{3}}{3!}x^4 + \cdots$

$(1-x)^{-2/3} = 1 + \dfrac{2}{3}x + \dfrac{2 \cdot 5}{3 \cdot 6} + \dfrac{2 \cdot 5 \cdot 8}{3 \cdot 6 \cdot 9}x^3 + \dfrac{2 \cdot 5 \cdot 8 \cdot 11}{3 \cdot 6 \cdot 9 \cdot 12}x^4 + \cdots$

Plug in $x = \dfrac{1}{2}$ to get:

$(1-\frac{1}{2})^{-2/3} = 1 + \dfrac{2}{3}\cdot\dfrac{1}{2} + \dfrac{2 \cdot 5}{3 \cdot 6}\cdot\dfrac{1}{2^2} + \dfrac{2 \cdot 5 \cdot 8}{3 \cdot 6 \cdot 9}\cdot\dfrac{1}{2^3} + \dfrac{2 \cdot 5 \cdot 8 \cdot 11}{3 \cdot 6 \cdot 9 \cdot 12}\cdot\dfrac{1}{2^4} + \cdots$

$2^{2/3} = 1 + \dfrac{2}{6} + \dfrac{2 \cdot 5}{6 \cdot 12}+ \dfrac{2 \cdot 5 \cdot 8}{6 \cdot 12 \cdot 18}+ \dfrac{2 \cdot 5 \cdot 8 \cdot 11}{6 \cdot 12 \cdot 18 \cdot 24} + \cdots$

Finally, subtract $1$ and divide both sides by $2$ to get:

$\dfrac{2^{2/3} - 1}{2} = \dfrac{1}{6} + \dfrac{5}{6 \cdot 12}+ \dfrac{ 5 \cdot 8}{6 \cdot 12 \cdot 18}+ \dfrac{ 5 \cdot 8 \cdot 11}{6 \cdot 12 \cdot 18 \cdot 24} + \cdots$

$\endgroup$
  • 1
    $\begingroup$ How have you determined the exponent to be $$-\frac23?$$ $\endgroup$ – lab bhattacharjee Sep 18 '14 at 8:21
  • $\begingroup$ I played around with different values until I found one that worked. I didn't really use any systematic method. $\endgroup$ – JimmyK4542 Sep 19 '14 at 1:20
1
$\begingroup$

$$\frac16+\frac{5}{6\cdot12}+\frac{5\cdot8}{6\cdot12\cdot18}+\frac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+\dots=$$

$$=\frac12\cdot\bigg[\frac26+\frac{2\cdot5}{6\cdot12}+\frac{2\cdot5\cdot8}{6\cdot12\cdot18}+\frac{2\cdot5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+\dots\bigg]=$$

$$=\frac12\cdot\bigg[\frac{(3-1)}{(6\cdot1)}+\frac{(3-1)\cdot(6-1)}{(6\cdot1)\cdot(6\cdot2)}+\frac{(3-1)\cdot(6-1)\cdot(9-1)}{(6\cdot1)\cdot(6\cdot2)\cdot(6\cdot3)}+\dots\bigg]=$$

$$=\frac12\cdot\bigg[\frac{(3\cdot1-1)}{(6\cdot1)}+\frac{(3\cdot1-1)\cdot(3\cdot2-1)}{(6\cdot1)\cdot(6\cdot2)}+\frac{(3\cdot1-1)\cdot(3\cdot2-1)\cdot(3\cdot3-1)}{(6\cdot1)\cdot(6\cdot2)\cdot(6\cdot3)}+\dots\bigg]=$$

$$=\frac12\cdot\sum_{n=1}^\infty\frac{\displaystyle\prod_{k=1}^n(3k-1)}{6^n\cdot n!}=\frac12\cdot\sum_{n=1}^\infty\frac{\displaystyle\prod_{k=1}^n\bigg(k-\frac13\bigg)}{2^n\cdot n!}=\frac12\cdot\sum_{n=1}^\infty\frac{\displaystyle\bigg(-\frac13\bigg){\Large!}\cdot\prod_{k=1}^n\bigg(k-\frac13\bigg)}{\bigg(-\dfrac13\bigg){\Large!}\cdot2^n\cdot n!}=$$

$$=\frac12\cdot\sum_{n=1}^\infty\frac{\displaystyle\bigg(n-\frac13\bigg){\Large!}}{\bigg(-\dfrac13\bigg){\Large!}\cdot2^n\cdot n!}=\frac12\cdot\sum_{n=1}^\infty\frac{\displaystyle\bigg(n-\frac13\bigg){\Large!}}{\bigg(-\dfrac13\bigg){\Large!}\cdot n!}\cdot\bigg(\frac12\bigg)^n=\frac12\cdot\sum_{n=1}^\infty{n-\frac13\choose n}\bigg(\frac12\bigg)^n$$

$$=\frac12\cdot\sum_{n=1}^\infty{\frac13-1\choose n}\bigg(-\frac12\bigg)^n=\frac12\cdot\bigg[-1+\sum_{n={\color{red}0}}^\infty{-\frac23\choose n}\bigg(-\frac12\bigg)^n\bigg]=$$

$$=\frac12\cdot\bigg[-1+\bigg(1-\frac12\bigg)^{^{-\tfrac23}}\bigg]=-\frac12+\frac1{\sqrt[3]2}.$$

$\endgroup$
  • $\begingroup$ We have used the fact that $\displaystyle{n-a\choose n}=(-1)^n{a-1\choose n}$. $\endgroup$ – Lucian Sep 18 '14 at 8:22
  • $\begingroup$ It goes on without saying that this is a binomial series. $\endgroup$ – Lucian Sep 18 '14 at 8:35
0
$\begingroup$

$$\begin{align} y &= \sum_{k=1} \frac{\prod_{j=2}^{k} 3j - 1} {\prod_{j=1}^k 6j} \tag{A} \\ &= \left(-\frac 12\right) + \sum_{k=0} \frac{\left(-\frac 12\right)\prod_{j=0}^{k} 3j - 1} {\prod_{j=1}^k 6j} \tag{B} \\ &= \left(-\frac 12\right) + \sum_{k=0} \frac{\left(-\frac 13\right)^{k+1}\left(-\frac 12\right)\prod_{j=0}^k 1/3 - j} {6^k k!}\tag{C} \\ &= \left(-\frac 12\right) + \left(-\frac 13\right)\left(-\frac 12\right)\sum_{k=0} \frac{\prod_{j=0}^k 1/3 - j} {k!}\left(-\frac 1 {3\cdot 6}\right)^k\tag{D} \\ &= \left(-\frac 12\right) + \left(\frac 16\right) \sum_{k=0} \frac{\prod_{j=0}^k 1/3 - j} {k!}\left(-\frac 1 {18}\right)^k\tag{E} \end{align}$$

Generalized binomial theorem is:

$$(1 + x)^n = \sum_{k=0} \frac{ \prod_{j=0}^k n - j } {k!} x^k\tag{F}$$

So if I don't have a type then

$$y = \left(-\frac 12\right) + \left(\frac 16\right)\left(1 + -\frac{1}{18}\right)^{1/3} = \frac{1}{18}\sqrt[3]{\frac{17}{2}} - \frac 12\tag{G}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.