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Find the sum of the series to infinity$$\frac{1}{4}+\frac{1\cdot 3}{4\cdot 6}+\frac{1\cdot 3\cdot 5}{4\cdot 6\cdot 8}+\cdots$$

Attempt-

I wrote the general term as $$\frac{\binom{2n}{n}}{2^{2n}\cdot (n+1)}$$ I don't know what to do next

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  • $\begingroup$ Does ' * ' refer to multiplication? $\endgroup$ Feb 25, 2015 at 8:48
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    $\begingroup$ I think that the formula you used for the general term is not correct. $\endgroup$ Feb 25, 2015 at 8:52

4 Answers 4

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The numerator of the $n$-th term is

$$\frac{(2n)!}{2^nn!}\;,$$

and the denominator is

$$\frac{2^{n+1}(n+1)!}2=2^n(n+1)!\;,$$

so the term is actually

$$\frac{(2n)!}{2^{2n}n!(n+1)!}=\frac{1}{n+1}\binom{2n}{n}\left(\frac{1}{4}\right)^n\;.$$

Do you know anything about the Catalan numbers and generating functions?

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  • $\begingroup$ I wonder if you did not miss a factor $2$ somewhere since the first terms look to be twice larger that those given in the post. Cheers :-) $\endgroup$ Feb 25, 2015 at 9:12
  • $\begingroup$ @Claude: $n$, $n+1$, who's counting? :-) Thanks! $\endgroup$ Feb 25, 2015 at 9:21
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It is not trivial, but your series is a telescopic one. Let $a_n=\frac{1}{4^n}\binom{2n}{n}$. We have:

$$ a_{n}-a_{n+1} = \frac{1}{4^n}\binom{2n}{n}-\frac{1}{4^{n+1}}\binom{2n+2}{n+1} =\frac{a_n}{2n+2}$$ so if we set $b_n = \frac{1}{(n+1)4^n}\binom{2n}{n}$ it follows that:

$$\sum_{n\geq 1}b_n = 2\sum_{n\geq 1}\frac{a_n}{2n+2} = 2\sum_{n\geq 1}\left(a_n-a_{n+1}\right) = 2a_1 = \color{red}{1}$$ since $\lim_{n\to +\infty}a_n = 0$.

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  • $\begingroup$ Nice Explanation Jack D'Aurizio. $\endgroup$
    – juantheron
    Jun 3, 2016 at 3:05
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$$S = 2\left[\frac{1}{2\cdot 4}+\frac{1\cdot 3}{2\cdot 4\cdot 6}+\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 8}+\cdots\right]$$

Here $$T_{n}\bf{(n^{th}) term } = 2\cdot \left[\frac{1\cdot 3 \cdot 5......(2n-1)}{2\cdot 4 \cdot 6.........(2n)(2n+2)}\right]$$

$$ = 2\left[\frac{1\cdot 3 \cdot 5......(2n-1)}{2\cdot 4 \cdot 6...(2n)(2n+2)}((2n+2)-(2n+1))\right]$$

So $$T_{n} = 2\left[\frac{1\cdot 3 \cdot 5......(2n-1)}{2\cdot 4 \cdot 6.........(2n)}-\frac{1\cdot 3 \cdot 5......(2n+1)}{2\cdot 4 \cdot 6.........(2n)}\right]$$

So $$S_{\infty} = 2\sum^{\infty}_{r=1}\left[\frac{1\cdot 3 \cdot 5......(2r-1)}{2\cdot 4 \cdot 6.........(2r)}-\frac{1\cdot 3 \cdot 5......(2r+1)}{2\cdot 4 \cdot 6.........(2r)}\right] = 2\cdot \frac{1}{2} = 1$$

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My attempt-

write down the Taylor series for $(1-x)^{\frac12}$ $$(1-x)^{\frac12}=1-\frac12x-\frac12\cdot\frac12\cdot\frac1{2!}x^2-\frac12\frac12\frac32\frac1{3!}x^3...$$ After putting x=1 and simplifying, I am getting value as 1.

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  • $\begingroup$ Note that I said the required sum is 1. $\endgroup$ Mar 3, 2015 at 3:22

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