# Simplifying an infinite sum

Question:

The sum of $$1-\frac16+\frac16\times\frac14-\frac16\times\frac14\times\frac{5}{18}+\cdots$$

is:

A) $$\frac23$$ B)$$\frac{2}{\sqrt3}$$ C)$$\sqrt\frac23$$ D)$$\frac{\sqrt3}{2}$$

After looking at the options I thought factoring into two's and three's would be a reasonable approach. The $$\frac{5}{18}$$ I considered factoring first as $$\frac{2+3}{2\times3^2}$$ and then as $$\frac{3^2-2^2}{2\times3^2}$$, but unfortunately I couldn't find a discernible pattern in either case. Any pointers in the right direction would be appreciated.

• Hint: $\frac14 = \frac{3}{12}$. The $n$-th term is $(-1)^n\frac{(2n-1)!!}{6^n n!} = \left(-\frac{1}{12}\right)^n \binom{2n}{n}$. $\binom{2n}{n}$ is known as the central binomial coefficient, look at its wiki entry and you will know how to compute the sum. – achille hui Nov 29 '18 at 9:43
• The pattern is completely unclear (to me, at least), but it does suggest an alternating sum of decreasing terms, from which we can infer that the answer is less than $1$ but greater than $1-{1\over6}={5\over6}=0.833333\ldots$. This rules out the first three options (e.g., $\sqrt{2/3}=0.816496\ldots$), leaving only $\sqrt3/2=0.866025\ldots$ as a possibly correct answer. – Barry Cipra Nov 29 '18 at 9:44
• Use find $$\left(1+\dfrac13\right)^{-1/2}$$ – lab bhattacharjee Nov 29 '18 at 9:52
• @s0ulr3aper07, In math.stackexchange.com/questions/746388/…, find en.wikipedia.org/wiki/Binomial_series#Convergence – lab bhattacharjee Nov 29 '18 at 10:35
• Also keep in mind efunda.com/math/exp_log/series_exp.cfm – lab bhattacharjee Nov 29 '18 at 10:38