Easy sum...

Calculus Level 3

lim ⁡ n → ∞ ∑ k = 0 n 1 ( n k ) = L \lim_{n \to \infty} \sum_{k=0}^n \frac 1{\binom nk}=L

For L L as defined above, find the value of L L L^L .

Notation: ( n k ) = n ! k ! ( n − k ) ! \dbinom nk =\dfrac{n!}{k!(n-k)!} denotes the binomial coefficient .


The answer is 4.

Francisco López by Francisco López , 21, Mexico

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1 solution

Naren Bhandari
Apr 2, 2020

We have the identity L = ∑ k = 0 n ( n k ) − 1 = n + 1 2 n + 1 ∑ k = 1 n + 1 2 k k L=\sum_{k=0}^n { n\choose k}^{-1} \ =\frac{n+1}{2^{n+1}}\sum_{k=1}^{n+1}\frac{2^{k}}{k} and hence we have lim ⁡ n → ∞ L = lim ⁡ n → ∞ n + 1 2 n + 1 ∑ k = 1 n + 1 2 k k n + 1 2 n + 1 ( 2 n + 2 n + 1 ) ⏟ S. C theorem = 2 \lim_{n\to\infty} L = \lim_{n\to\infty}\frac{n+1}{2^{n+1}}\sum_{k=1}^{n+1} \frac{2^k}{k} \\ \frac{n+1}{2^{n+1}}\underbrace{\left(\frac{2^{n+2}}{n+1}\right)}_{\text{S. C theorem }}=2 and thus we have L L = 4 L^L=4

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