Find sum of series $\frac{1}{6} + \frac{5}{6 \cdot 12} + \frac{5\cdot 8}{6 \cdot 12 \cdot 18} + \dotsb$

Asked Aug 11 '17 at 11:29

Active Aug 11 '17 at 13:06

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Find the sum of the following infinite series: $$ \frac{1}{6} + \frac{5}{6 \cdot 12} + \frac{5\cdot 8}{6 \cdot 12 \cdot 18} + \frac{5\cdot 8 \cdot 11}{6 \cdot 12 \cdot 18 \cdot 24} + \dotsb$$

I was trying this question many times but I could not get it. I was trying to solve this problem by D'Alembert ratio test, but I did not get my answer. My teacher gave me this question as a homework, I don't any hint or idea to solved this question, I was trying many times but didn't get it.

If anybody help me I would be very thankful. Thank you.

edited Aug 11 '17 at 11:40

SvanN

2,307

asked Aug 11 '17 at 11:29

jasmine

14,457

D'Alembert is used to find the radius of convergence, and tells nothing about the value of a convergent series. – Vim Aug 11 '17 at 11:34
Looks like a case of the binomial theorem. – Angina Seng Aug 11 '17 at 11:48
It seems that it's like $-\sqrt[3]{1-x}$ – Michael Rozenberg Aug 11 '17 at 11:54
The first term of $\frac16$ does not really fit the others – Henry Aug 11 '17 at 12:00
$$\sum_{n=1}^\infty\prod_{k=1}^n\dfrac{1}{2^{k+1}}\left(1-\dfrac{1}{3^k}\right)$$ – Nosrati Aug 11 '17 at 12:41
The generating function is $g(z)=\dfrac{(2-z)^{2/3}-2^{2/3}}{2 (2-z)^{2/3} z}$ and the sum is $g(1)=\frac{1}{2}-\frac{1}{\sqrt[3]{2}}$ – Raffaele Aug 11 '17 at 13:27
my teacher said that answer 1/2(4^1/3 -1) i don't know how it come @ raffaele – jasmine Aug 11 '17 at 13:34

Find sum of series $\frac{1}{6} + \frac{5}{6 \cdot 12} + \frac{5\cdot 8}{6 \cdot 12 \cdot 18} + \dotsb$

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