I need help on finding the $n^{\text{th}}$ term of this infinite series? $$ s=\frac{1}{4}+\frac{1\cdot 3}{4\cdot 6}+\frac{1\cdot 3\cdot 5}{4\cdot 6\cdot 8}+\ldots $$ Could you help me in writing the general term/solving?

  • $\begingroup$ What do you think the next term ought to be? What about the $k$th term? $\endgroup$ – Eric Tressler Aug 10 '13 at 13:31
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    $\begingroup$ How about using LATEX? Everyone we'll understand you better, because now I don't get the expression $\endgroup$ – Stefan4024 Aug 10 '13 at 13:39
  • $\begingroup$ I have written up the formula as I understood it; is this what you intended? Please, consider looking at how I've modified your post so that you can do this yourself going forward. $\endgroup$ – Nick Peterson Aug 10 '13 at 13:50
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    $\begingroup$ Consider the numerator: $1\times3\times5\times7 = \frac{1\times2\times3\times4\times5\times6\times7}{2\times4\times6} = \frac{1\times2\times3\times4\times5\times6\times7}{2^3(1\times2\times3)} = \frac{7!}{2^3 3!}$. Can you generalize this? Can you do something similar for the denominator? $\endgroup$ – TenaliRaman Aug 10 '13 at 13:53
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    $\begingroup$ @TenaliRaman It's a little cleaner to write it as $\frac{8!}{2^44!}$. Then we more easily see the pattern. Basically, add an $8$ to the numerator and denominator in the first step. $\endgroup$ – Thomas Andrews Aug 10 '13 at 14:20

The first thing you can do is start with $a_1=\frac{1}{4}$ and then realize that $$a_{n+1} =a_n\frac{2n-1}{2n+2}$$ that doesn't seem to get you anywhere, however.

As commentator Tenali notes, you can write the numerator as $$1\cdot 3\cdots (2n-1) = \frac{1\cdot 2 \cdot 3 \cdots (2n)}{2\cdot 4\cdots (2n)}=\frac{(2n)!}{2^nn!}$$

The denominator, on the other hand, is $$ 4\cdot 6\cdots \left(2(n+1)\right) = 2^n\left(2\cdot 3\cdots (n+1)\right) = 2^n(n+1)!$$

So this gives the result:

$$a_n = \frac{(2n)!}{2^{2n} n!(n+1)!} = \frac{1}{2^{2n}}\frac{1}{n+1}\binom{2n}{n} = \frac{1}{2^{2n}}C_n$$

where $C_n$ is the $n^{\text{th}}$ Catalan number.

If all you want is then $n^{\text{tn}}$ term, that might be enough - you can even skip the part about Catalan numbers and just write it as $a_n=\frac{1}{4^n(n+1)}\binom{2n}n$.

As it turns out, the Catalan numbers have a generating function (see the link above:)

$$\frac{2}{1+\sqrt{1-4x}} = \sum_{n=0}^\infty C_nx^n$$

So, if the series converges when $x=\frac{1}{4}$, it converges to $2$.

(It does converge, since $C_n \sim \frac{2^{2n}}{n^{3/2}\sqrt{\pi}}$.)


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