It should be in Combinatorics section

Calculus Level 5

$\large \displaystyle \lim _{ n\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ \frac { ^{ 2r }{ { C }_{ r } } }{ { 2 }^{ 2r }\left( 2r+1 \right) } } } = \frac { \pi }{ a } -b$

If the series above is true for constants $a,b$ , find the value of $a+b$ .

This question is part of set Serieses are fun! .

7 3 5 2 1 8 6 4

2 solutions

Chew-Seong Cheong
Dec 8, 2015

Just showing what Abhishek Sharma meant.

$\begin{aligned} \sum_{r=1}^\infty \frac{^{2r}C_r}{2^{2r}(2r+1)} & = \frac{2}{4\dot{}3} + \frac{6}{16\dot{}5} + \frac{20}{64\dot{}7} + \frac{70}{256\dot{}9} +... \\ & = \frac{1}{6} + \frac{3}{40} + \frac{5}{112} + \frac{35}{1152} +... \quad \quad \small \color{#3D99F6}{\text{As }\sin^{-1} x = x + \frac{1}{6}x^3 + \frac{3}{40}x^5 + \frac{5}{112}x^7 + \frac{35}{1152}x^9 +... } \\ & = \sin^{-1} (1) - 1 = \frac{\pi}{2} - 1 \end{aligned}$

$\Rightarrow a+b = 2 + 1 = \boxed{3}$

Incredible Mind
May 19, 2015

this may be lengthy..but just use this equation..

substitute this into the sum.

interchange sum and integral since bothr sums

.first do the sum then the integral.

make 2^2r = 4^r

sum is simple by using infinite GP,|x|<1..becoz integral from 0 to 4.. some diff and integrals required in this but u can bcoz i is elementary

ALL THIS COMES FORM CONCEPT OF CATALAN NUBERS

ANS is

pi/2 - 1

Alternative way

Simplify a bit and use the maclaurin series of $\arcsin { x }$ .

Abhishek Sharma - 6 years ago

very nice..it is just arcsin1 - 1

i guess this doesnt deserve to be level 5 bcoz it is so direct

incredible mind - 6 years ago

Yes, I agree that it shouldn't be Level 5 but should be Level 4.

Not many people could do it because maclaurin series of $\arcsin { x }$ is not that common. Also because it has binomial coefficient it looks a bit scary.

Abhishek Sharma - 6 years ago

It should be in Combinatorics section

This question is part of set Serieses are fun! .

2 solutions

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