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The sum of $n$ terms of the series $$\frac{1}{2}+\frac{1}{2!}\left(\frac{1}{2}\right)^2+\frac{1\cdot 3}{3!}\left(\frac{1}{2}\right)^3+\frac{1\cdot 3 \cdot 5}{4!}\left(\frac{1}{2}\right)^4+\frac{1\cdot 3 \cdot 5 \cdot 7}{5!}\left(\frac{1}{2}\right)^5+....$$

And also calculate sum of $\infty$ terms.

$\bf{My\; Try::}$ We can write above series as $$ 1\underbrace{-\frac{1}{2}+\frac{1}{2!}\left(\frac{1}{2}\right)^2+\frac{1\cdot 3}{3!}\left(\frac{1}{2}\right)^3+\frac{1\cdot 3 \cdot 5}{4!}\left(\frac{1}{2}\right)^4+\frac{1\cdot 3 \cdot 5 \cdot 7}{5!}\left(\frac{1}{2}\right)^5+....}_{S_{n}}$$

So here $$\bf{r^{th}}\; terms \; of \; above\; series (T_{r}) = \frac{1\cdot 3 \cdot 5\cdot \cdot \cdot \cdot (2r-3)}{r!}\cdot \frac{1}{2^r}$$

So $$T_{r} = \frac{1}{3}\left[\frac{1\cdot 3 \cdot 5\cdot \cdot \cdot \cdot (2r-5)}{(r-1)!}\cdot \frac{1}{2^{r-1}}-\frac{1\cdot 3 \cdot 5\cdot \cdot \cdot \cdot (2r-3)}{r!}\cdot \frac{1}{2^r}\right]$$

So $$S_{n} = \sum^{n}_{r=1}T_{r} = \frac{1}{3}\sum^{n}_{r=1}\left[\frac{1\cdot 3 \cdot 5\cdot \cdot \cdot \cdot (2r-5)}{(r-1)!}\cdot \frac{1}{2^{r-1}}-\frac{1\cdot 3 \cdot 5\cdot \cdot \cdot \cdot (2r-3)}{r!}\cdot \frac{1}{2^r}\right]$$

So $$S_{n} = \frac{1}{3}\left[-3-\frac{1\cdot 3 \cdot 5\cdot \cdot \cdot \cdot (2n-3)}{n!}\cdot \frac{1}{2^n}\right] = -1+\frac{1}{3}\frac{1\cdot 3 \cdot 5\cdot \cdot \cdot \cdot (2n-3)}{n!}\cdot \frac{1}{2^n}$$

So our Sum is $$=\frac{1}{3}\frac{1\cdot 3 \cdot 5\cdot \cdot \cdot \cdot (2n-3)}{n!}\cdot \frac{1}{2^n}$$

Is my process is right or not, If not the how can i calculate it, Thanks

Community
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juantheron
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1 Answers1

7

This series looks like \begin{align*} \frac{1}{2}&+\sum_{r=2}^\infty\frac{1}{r!}\left(\frac{1}{2}\right)^r(2r-3)!!\tag{1}\\ &=\frac{1}{2}+\sum_{r=1}^\infty\frac{1}{(r+1)!}\left(\frac{1}{2}\right)^{r+1}(2r-1)!!\\ &=\frac{1}{2}+\sum_{r=1}^\infty\frac{1}{(r+1)!}\left(\frac{1}{2}\right)^{r+1}\frac{(2r)!}{(2r)!!}\tag{2}\\ &=\frac{1}{2}+\sum_{r=1}^\infty\frac{1}{(r+1)!}\left(\frac{1}{2}\right)^{r+1}\frac{(2r)!}{2^rr!}\tag{3}\\ &=\frac{1}{2}+\frac{1}{2}\sum_{r=1}^\infty\frac{1}{r+1}\binom{2r}{r}\left(\frac{1}{4}\right)^r\tag{4}\\ &=\frac{1}{2}\sum_{r=0}^\infty\frac{1}{r+1}\binom{2r}{r}\left(\frac{1}{4}\right)^r\\ &=\frac{1}{2}\left.\left(\frac{1-\sqrt{1-4x}}{2x}\right)\right|_{x=\frac{1}{4}}\tag{5}\\ &=\frac{1}{2}\cdot 2\\ &=1 \end{align*}

Comment:

  • In (1) we use double factorial notation $(2r-3)!!=(2r-3)\cdot(2r-5)\cdots 5\cdot 3\cdot 1$

  • In (2) we use $r!=r!!\cdot(r-1)!!$

  • In (3) we use $(2r)!!=2^r r!$

  • In (4) we obtain the Catalan numbers $C_r=\frac{1}{r+1}\binom{2r}{r}$

  • In (5) we use the generating function of the Catalan numbers

Markus Scheuer
  • 108,315
  • Nicely done! Once one recognizes the Catalan numbers, getting the final result is simple enough. The one thing I'm left wondering is if there's some probabilistic interpretation, i.e. $p_r = 2^{-2r-1} C_r$ as the probability of some $r$th outcome. – Semiclassical Aug 09 '16 at 19:27
  • Upon reflection, I think the following works: I flip a coin repeatedly until heads and tails have come up an equal number of times. What's the probability that $2r$ such flips are required? – Semiclassical Aug 09 '16 at 19:33
  • @Semiclassical: Thanks for your nice comment and really good reflection! :-) It is e.g. stated in terms of paths as theorem in section 9, chapter III of this classic. – Markus Scheuer Aug 09 '16 at 19:55