It's Always e!

Calculus Level 4

lim ⁡ n → ∞ n − n 2 ( ( n + 1 ) ( n + 1 2 ) ( n + 1 2 2 ) ⋯ ( n + 1 2 n − 1 ) ) n = e k \lim_{n\to\infty} n^{-n^{2}} \left((n+1)\left(n+\frac{1}{2}\right)\left(n+\frac{1}{2^{2}}\right)\cdots\left(n+\frac{1}{2^{n-1}}\right)\right)^{n} = e^{k}

Find the value of k k .


Notation: e ≈ 2.71828 e \approx 2.71828 denotes the Euler's number .


The answer is 2.

Thomas Jacob by Thomas Jacob , 19, India

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1 solution

Chew-Seong Cheong
Oct 25, 2017

L = lim ⁡ n → ∞ n − n 2 ( ( n + 1 ) ( n + 1 2 ) ( n + 1 2 2 ) ( n + 1 2 n − 1 ) ) n = lim ⁡ n → ∞ ( ∏ i = 0 n − 1 ( n + 1 2 i ) n n ) n = lim ⁡ n → ∞ ∏ i = 0 n − 1 ( 1 + 1 2 i n ) n = lim ⁡ n → ∞ ∏ i = 0 n − 1 ( 1 + 1 2 i n ) 2 i n ⋅ 2 − i = lim ⁡ n → ∞ ∏ i = 0 n − 1 exp ⁡ ( 1 2 i ) where exp ⁡ ( x ) = e x = lim ⁡ n → ∞ exp ⁡ ∑ i = 0 n − 1 1 2 i = e 2 \begin{aligned} L & = \lim_{n \to \infty} n^{-n^2} \left(\left(n+1\right)\left(n+\frac 12\right)\left(n+\frac 1{2^2}\right)\left(n+\frac 1{2^{n-1}}\right)\right)^n \\ & = \lim_{n \to \infty} \left(\frac {\prod_{i=0}^{n-1}\left(n+\frac 1{2^i}\right)}{n^n} \right)^n \\ & = \lim_{n \to \infty} \prod_{i=0}^{n-1}\left(1+\frac 1{2^in}\right)^n \\ & = \lim_{n \to \infty} \prod_{i=0}^{n-1}\left(1+\frac 1{2^in}\right)^{2^in\cdot 2^{-i}} \\ & = \lim_{n \to \infty} \prod_{i=0}^{n-1} \exp \left(\frac 1{2^i}\right) & \small \color{#3D99F6} \text{where }\exp(x) = e^x \\ & = \lim_{n \to \infty} \exp \sum_{i=0}^{n-1} \frac 1{2^i} \\ & = e^2 \end{aligned}

⟹ k = 2 \implies k = \boxed{2}

7 Helpful 0 Interesting 0 Brilliant 1 Confused

Nice problem and solution.

Aman Joshi - 3 years, 7 months ago

There is typo in second step 2 k 2^k where i = 0 i=0

Naren Bhandari - 3 years, 7 months ago

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Thanks. I have changed it. After keying the solution with k k all over the place. Then I realised the answer requested for is k k , so I replace the k k with i i and missed one.

Chew-Seong Cheong - 3 years, 7 months ago

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