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What is the sum of the series

$$\frac {4}{10}+\frac {4\cdot7}{10\cdot20}+ \frac {4\cdot7\cdot10}{10\cdot20\cdot30}+\cdots?$$

I know how to check if a series is convergent or not.Is there any technique to find out the sum of a series like this where each term increases by a pattern?

GNUSupporter 8964民主女神 地下教會
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jimm
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    In your title, it seems you have base ten numbers. In the body of the question, I would guess 4.7.10 means multiply, giving 280. Anyway, typesetting needs considerable work – Will Jagy Jan 18 '18 at 02:45
  • Please check it now. – jimm Jan 18 '18 at 02:51
  • better, and now consistent – Will Jagy Jan 18 '18 at 02:56
  • Easy to see sum lies between 3/7 and 2/3. Can't find value – King Tut Jan 18 '18 at 03:12
  • @Manthanein No, it isn't. Note the difference in the denominator. This requires extra care is the linked answer. Compare and contrast my solution with the one in another post, especially in terms of the number of lines of the solution. – GNUSupporter 8964民主女神 地下教會 Jan 18 '18 at 03:23
  • @GNU Supporter The OP might get many different ideas from that question. Moreover he/she might intend to try on his/her own to solve the problem instead of just opening a thread on account of just the difference in the denominator. Trying on his/her own might increase the critical thinking skill too. So I flagged I as duplicate. If it doesn't fit in the norms of duplicate I will certainly try to unflag it. – Rohan Shinde Jan 18 '18 at 03:31
  • type sum[product[(1+3k)/(10k), {k,1,n}],{n,1,infinity}] into Wolfram alpha – Steven Alexis Gregory Jan 18 '18 at 03:39

1 Answers1

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My solution is adapted from another similar solution.

\begin{align} \sum_{n=1}^{\infty}\frac{4\cdot7\cdot\cdots\cdot(3n+1)}{n!10^n} &=\sum_{n=1}^{\infty}\frac{\frac43\cdot\frac73\cdot\cdots\cdot\frac{3n+1}3}{n!\left(10/3\right)^n}\\ &=\sum_{n=1}^{\infty}\binom{\frac{3n+1}{3}}{n}\left(\frac{3}{10}\right)^n\\ &=\left[\sum_{n=1}^{\infty}\binom{\frac{3n+1}{3}}{n}x^n\right]_{x=3/10}\\ &=\left[\sum_{n=1}^{\infty}\binom{-\frac{4}{3}}{n}(-x)^{n}\right]_{x=3/10}\\ &=\left[ \left(\left(1-x\right)^{-4/3}-1\right) \right]_{x=3/10}\\ &=\sqrt[3]{\left(\frac{10}{7}\right)^4}-1 \end{align}

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