A Weird Little Series...

Algebra Level 5

1 + ∑ m = 1 ∞ ( − 1 ) m P odd ( m ) 2 m m ! \large1+\sum_{m=1}^{\infty} (-1)^{m}\frac{P_{\text{odd}}(m)}{2^{m}m!}

The above expression can be written as n , \sqrt{n}, where n n is a positive integer.

What is n ? n?

Note: P odd ( m ) = ∏ k = 0 m − 1 ( 2 k − 1 ) P_{\text{odd}}(m)=\prod\limits_{k=0}^{m-1}(2k-1)


The answer is 2.

Andrei Li by Andrei Li , 16, Canada

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1 solution

Chew-Seong Cheong
Oct 16, 2018

Consider the binomial expansion of 1 + x \sqrt{1+x} as follows:

1 + x = 1 + 1 2 x − 1 2 ⋅ 1 2 ⋅ x 2 2 ! + 1 2 ⋅ 1 2 ⋅ 3 2 ⋅ x 3 3 ! − 1 2 ⋅ 1 2 ⋅ 3 2 ⋅ 5 2 ⋅ x 4 4 ! + 1 2 ⋅ 1 2 ⋅ 3 2 ⋅ 5 2 ⋅ 7 2 ⋅ x 5 5 ! − ⋯ = 1 + ∑ m = 1 ( − 1 ) m ∏ k = 0 m − 1 ( 2 k − 1 ) 2 m m ! x m Putting x = 1 2 = 1 + ∑ m = 1 ( − 1 ) m ∏ k = 0 m − 1 ( 2 k − 1 ) 2 m m ! \begin{aligned} \sqrt{1+x} & = 1 + \frac 12 x - \frac 12 \cdot \frac 12 \cdot \frac {x^2}{2!} + \frac 12 \cdot \frac 12 \cdot \frac 32 \cdot \frac {x^3}{3!} - \frac 12 \cdot \frac 12 \cdot \frac 32 \cdot \frac 52 \cdot \frac {x^4}{4!} + \frac 12 \cdot \frac 12 \cdot \frac 32 \cdot \frac 52 \cdot \frac 72 \cdot \frac {x^5}{5!} - \cdots \\ & = 1 + \sum_{m=1}(-1)^m \frac {\prod_{k=0}^{m-1}(2k-1)}{2^m m!}x^m \quad \quad \small \color{#3D99F6} \text{Putting }x = 1 \\ \sqrt 2 & = 1 + \sum_{m=1}(-1)^m \frac {\prod_{k=0}^{m-1}(2k-1)}{2^m m!} \end{aligned}

Therefore, n = 2 n = \boxed 2 .

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