Relating products and taylor series

Calculus Level 5

∏ m = 1 ∞ [ 1 2 exp ⁡ ( ∑ n = 1 m ( − 1 ) n − 1 n ) ] = exp ⁡ ( 1 ) 2 × exp ⁡ ( 1 − 1 2 ) 2 × exp ⁡ ( 1 − 1 2 + 1 3 ) 2 × exp ⁡ ( 1 − 1 2 + 1 3 − 1 4 ) 2 × ⋯ \prod_{m=1}^\infty \left [ \dfrac12 \exp \left(\sum_{n=1}^m \dfrac{ (-1)^{n-1}}n \right) \right ] = \dfrac{\exp(1)}2 \times \dfrac{\exp\left (1- \frac12\right)}2 \times\dfrac{\exp\left (1- \frac12+ \frac13\right)}2 \times\dfrac{\exp\left (1- \frac12+ \frac13- \frac14\right)}2 \times \cdots

Evaluate the infinite product above, where exp ⁡ ( x ) = e x \exp(x) = e^x .

If your answer can be expressed as r × e s , r \times e^{s}, where r r and s s are rational numbers, give your answer as r + s r+s .

Bonus: Give a closed form of ∏ m = 1 ∞ [ 1 1 − x exp ⁡ ( − ∑ n = 1 m x n n ) ] \displaystyle \prod_{m=1}^\infty \left [ \dfrac1{1-x} \exp \left(-\sum_{n=1}^m \dfrac{x^{n}}n \right) \right ] for − 1 ≤ x < 1 -1\leq x < 1 .


The answer is 1.5.

Nicolas Nagel by Nicolas Nagel , 22, Germany

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1 solution

Mark Hennings
Sep 16, 2017

Since ∏ m = 1 2 M [ 1 2 exp ⁡ ( ∑ n = 1 m ( − 1 ) n − 1 n ) ] = 1 2 2 M exp ⁡ [ ∑ m = 1 2 M ∑ n = 1 m ( − 1 ) n − 1 n ] = 1 2 2 M exp ⁡ [ ∑ n = 1 2 M ( − 1 ) n − 1 ( 2 M + 1 − n ) n ] = 1 2 2 M exp ⁡ [ ( 2 M + 1 ) ∑ n = 1 2 M ( − 1 ) n − 1 n ] = exp ⁡ [ ( 2 M + 1 ) ( H 2 M − H M ) − 2 M ln ⁡ 2 ] \begin{aligned} \prod_{m=1}^{2M}\left[ \frac12 \exp\left(\sum_{n=1}^m \frac{(-1)^{n-1}}{n}\right)\right] & = \; \frac{1}{2^{2M}}\exp\left[\sum_{m=1}^{2M}\sum_{n=1}^m \frac{(-1)^{n-1}}{n}\right] \; = \; \frac{1}{2^{2M}} \exp\left[\sum_{n=1}^{2M}\frac{(-1)^{n-1}(2M+1-n)}{n}\right] \\ & = \; \frac{1}{2^{2M}}\exp\left[(2M+1)\sum_{n=1}^{2M}\frac{(-1)^{n-1}}{n}\right] \; = \; \exp\big[(2M+1)(H_{2M} - H_M) - 2M\ln2\big] \end{aligned} Since H N = ln ⁡ N + γ + 1 2 N + O ( N − 2 ) N → ∞ H_N \; = \; \ln N + \gamma + \frac{1}{2N} + O\big(N^{-2}\big) \hspace{2cm} N \to \infty we see that ( 2 M + 1 ) ( H 2 M − H M ) − 2 M ln ⁡ 2 = ln ⁡ 2 − 2 M + 1 4 M + O ( N − 1 ) N → ∞ (2M+1)(H_{2M} - H_M) - 2M\ln 2 \; = \; \ln2 - \frac{2M+1}{4M} + O\big(N^{-1}\big) \hspace{2cm} N \to \infty and hence, letting M → ∞ M \to \infty , ∏ m = 1 ∞ [ 1 2 exp ⁡ ( ∑ n = 1 m ( − 1 ) n − 1 n ) ] = exp ⁡ [ ln ⁡ 2 − 1 2 ] = 2 e − 1 2 \prod_{m=1}^{\infty}\left[ \frac12 \exp\left(\sum_{n=1}^m \frac{(-1)^{n-1}}{n}\right)\right] \; = \; \exp\big[\ln2 - \tfrac12\big] \; = \; 2e^{-\frac12} making the answer 2 − 1 2 = 3 2 2 - \tfrac12 = \boxed{\tfrac32} .

The result for − 1 < x < 1 -1 < x < 1 is perhaps a little easier. ∏ m = 1 M [ 1 1 − x exp ⁡ ( − ∑ n = 1 m x n n ) ] = 1 ( 1 − x ) M exp ⁡ [ − ∑ m = 1 M ∑ n = 1 m x n n ] = 1 ( 1 − x ) M exp ⁡ [ − ∑ n = 1 M ( M + 1 − n ) x n n ] = 1 ( 1 − x ) M exp ⁡ [ x ( 1 − x M ) 1 − x − ( M + 1 ) ∑ n = 1 M x n n ] = exp ⁡ [ x ( 1 − x M ) 1 − x − ( M + 1 ) ∑ n = 1 M x n n − M ln ⁡ ( 1 − x ) ] = exp ⁡ [ x ( 1 − x M ) 1 − x − ∑ n = 1 M x n n + M ∑ n = M + 1 ∞ x n n ] \begin{aligned} \prod_{m=1}^M \left[\frac{1}{1-x}\exp\left(-\sum_{n=1}^m \frac{x^n}{n}\right)\right] & = \; \frac{1}{(1-x)^M}\exp\left[-\sum_{m=1}^M\sum_{n=1}^m \frac{x^n}{n}\right] \; = \; \frac{1}{(1-x)^M}\exp\left[-\sum_{n=1}^M \frac{(M+1-n)x^n}{n} \right] \\ & = \; \frac{1}{(1-x)^M}\exp\left[\frac{x(1-x^M)}{1-x} - (M+1)\sum_{n=1}^M \frac{x^n}{n}\right] \; = \; \exp\left[ \frac{x(1-x^M)}{1-x} - (M+1)\sum_{n=1}^M \frac{x^n}{n} - M\ln(1-x)\right] \\ & = \; \exp\left[\frac{x(1-x^M)}{1-x} - \sum_{n=1}^M \frac{x^n}{n} + M\sum_{n=M+1}^\infty \frac{x^n}{n}\right] \end{aligned} Since ∣ M ∑ m = M + 1 ∞ x n n ∣ ≤ M M + 1 ∑ n = M + 1 ∞ ∣ x ∣ n ≤ ∣ x ∣ M + 1 1 − ∣ x ∣ \left| M\sum_{m=M+1}^\infty \frac{x^n}{n}\right| \; \le \; \frac{M}{M+1}\sum_{n=M+1}^\infty |x|^n \; \le \; \frac{|x|^{M+1}}{1-|x|} we deduce that lim ⁡ M → ∞ ( x ( 1 − x M ) 1 − x − ∑ n = 1 M x n n + M ∑ n = M + 1 ∞ x n n ) = x 1 − x + ln ⁡ ( 1 − x ) \lim_{M \to \infty} \left(\frac{x(1-x^M)}{1-x} - \sum_{n=1}^M \frac{x^n}{n} + M\sum_{n=M+1}^\infty \frac{x^n}{n}\right) \; = \; \frac{x}{1-x} + \ln(1-x) and hence ∏ m = 1 ∞ [ 1 1 − x exp ⁡ ( − ∑ n = 1 m x n n ) ] = ( 1 − x ) exp ⁡ [ x 1 − x ] ∣ x ∣ < 1 \prod_{m=1}^\infty \left[\frac{1}{1-x}\exp\left(-\sum_{n=1}^m \frac{x^n}{n}\right)\right] \; = \; (1-x)\exp\left[\frac{x}{1-x}\right] \hspace{2cm} |x| < 1

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