0
$\begingroup$

I have the following series:

$$1 + \frac{2}{3}\cdot\frac{1}{2} + \frac{2\cdot5}{3\cdot6}\cdot\frac{1}{2^2} + \frac{2\cdot5\cdot8}{3\cdot6\cdot9}\cdot\frac{1}{2^3} + \ldots$$

I have to find the value of this series, and I have four options:
(A) $2^{1/3}$ (B) $2^{2/3}$ (C) $3^{1/2}$ (D) $3^{3/2}$

I can't seem to find a general term for this. I tried:

$$S = 1 + \frac{(1 - \frac{1}{3})}{1!}(\frac{1}{2}) + \frac{(1 - \frac{1}{3})(2 - \frac{1}{3})}{2!}(\frac{1}{2})^2 + \frac{(1 - \frac{1}{3})(2 - \frac{1}{3})(3 - \frac{1}{3})}{3!}(\frac{1}{2})^3 + \ldots$$

But this doesn't seem to get me anywhere.

Any help?


This maybe a telescopic series, because there was a similar question we solved in class which ended up being telescopic:

$$ \frac{3}{2^3} + \frac{4}{2^4\cdot3} + \frac{5}{2^6\cdot3} + \frac{6}{2^7\cdot5} + \ldots$$

$=\displaystyle\sum\limits_{r=1}^\infty\frac{r+2}{2^{r+1}r(r+1)}$

$=\displaystyle\sum \bigg(\frac{1}{2^r r} - \frac{1}{2^{r+1}(r+1)}\bigg) = \frac{1}{2}$

$P.S:$ This problem was included in my set of questions for Binomial Theorem, which is why I thought it might be related to it.

$\endgroup$
4
  • $\begingroup$ $$\sum_{n=0}^\infty\frac{(3n-1)!!!}{(3n)!!!}2^{-n}=?$$ And it's not related to Newton's binomial. $\endgroup$ – Lucian Nov 5 '13 at 5:17
  • $\begingroup$ @Lucian What does the triple $!$ mean? Is it a nested factorial? And I've edited my question too :) $\endgroup$ – mikhailcazi Nov 5 '13 at 6:02
  • $\begingroup$ Just like the double factorial; only that it's triple. :-) $\endgroup$ – Lucian Nov 5 '13 at 6:05
  • $\begingroup$ It can't be telescopic, can it, since $\sqrt[3]4$ is not a rational quantity. $\endgroup$ – Lucian Nov 5 '13 at 6:21
2
$\begingroup$

You're looking for $$ 1+\sum_{k=1}^\infty \frac{\prod_{i=1}^k (3i-1)}{3^k k!}2^{-k} $$ That looks an awful lot like a Maclaurin series to me. What sort of a function would give us a Maclaurin series like this? I started with $f(x)=x^{-2/3}$, since continued derivation should give coefficients similar to those seen in the sum. Obviously we don't want to find the series around $0$, so we'll center the function around 1: $f(x-1)=(x-1)^{-2/3}$. If you compute the Maclaurin series, you obtain $$ 1+\sum_{k=1}^\infty \frac{\prod_{i=1}^k(3i-1)}{3^k}\frac{x^k}{k!}=(x-1)^{-2/3} $$ So $(\frac{1}{2}-1)^{-2/3}=\sqrt[3]{4}$ is your sum.

$\endgroup$
5
  • $\begingroup$ If you'd be so kind, could you tell me what Taylor-Maclaurin series yields the following result : $$1-5\cdot\left(\frac12\right)^3+9\cdot\left(\frac{1\cdot3}{2\cdot4}\right)^3- ... =\frac2\pi$$ $\endgroup$ – Lucian Nov 5 '13 at 7:11
  • $\begingroup$ I think I'd need to see more terms of that sequence, but it looks like $1+\sum (-1)^n(4n+1)(\frac{(2n)!}{4^n n!^2})^3 $, which I think would correspond to some sort of root, since the series for sqrt(x+1) has a similar sort of pattern (without the cubes... those look problematic). $\endgroup$ – Tim Ratigan Nov 5 '13 at 18:36
  • $\begingroup$ It's from Ramanujan's Second Notebook. There aren't any more explicit terms, but we're obviously dealing with $$\sum_{n=0}^\infty(-1)^n\cdot(4n+1)\cdot\left[\frac{(2n-1)!!}{(2n)!!}\right]^3=\frac2\pi$$ $\endgroup$ – Lucian Nov 5 '13 at 21:34
  • $\begingroup$ How do I go about computing the Maclaurin series? $\endgroup$ – mikhailcazi Nov 10 '13 at 6:30
  • $\begingroup$ That sounds like a question you may want to put on the forum, but the general idea of a Maclaurin series is $$ f(x-n)=\sum_{k=0}^\infty f^{(k)}(n)\cdot\frac{x^k}{k!} $$ Where $f^{(k)}(n)$ is the $k^{\mathrm{th}}$ derivative of $f$ at $n$. If you have an infinite sum of the terms of a Maclaurin series, then you can just replace it with the original function and evaluate that function. $\endgroup$ – Tim Ratigan Nov 10 '13 at 8:15
0
$\begingroup$

According to Mathematica, the sum is $2^{2/3}$. I don't know how to ask Mathematica to explain its method for deriving that result.

Define pr[n_] := Product[(3 i - 1)/(6 i), {i, 1, n}] Compute Sum[pr[n], {n, 0, [Infinity]}] Result: 2^(2/3)

$\endgroup$

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.