A probability problem by Tanishq Varshney

Probability Level 5

A = lim ⁡ n → ∞ ( ∑ r = 0 n ( n r ) n r ( r + 3 ) ) \large{A=\displaystyle \lim_{n\to \infty}(\sum^{n}_{r=0} \frac{\binom{n}{r}}{n^{r}(r+3)})}

Which of the following is correct

P ) P) A A is an irrational number

Q ) Q) l n A − 2 lnA-2 is rational

R ) R) A is a rational number

S ) S) l o g 2 A + 2 log_{2} A+2 is rational

PQ RS P S None QR

Tanishq Varshney by Tanishq Varshney , 23, India

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1 solution

A = lim ⁡ n → ∞ ∑ r = 0 n ( r n ) n r ( r + 3 ) = lim ⁡ n → ∞ ∑ r = 0 n n ! r ! ( n − r ) ! n r ( r + 3 ) = lim ⁡ n → ∞ ∑ r = 0 n [ lim ⁡ n → ∞ ( n ! n r ( n − r ) ! ) ( 1 r ! ( r + 3 ) ) ] = lim ⁡ n → ∞ ∑ r = 0 n 1 r ! ( r + 3 ) Note: lim ⁡ n → ∞ ( n ! n r ( n − r ) ! ) = 1 = lim ⁡ n → ∞ ∑ r = 0 n x r + 3 r ! ( r + 3 ) for x = 1 = lim ⁡ n → ∞ ∑ r = 0 n ∫ 0 1 x r + 2 r ! d x = ∫ 0 1 x 2 lim ⁡ n → ∞ ∑ r = 0 n x r r ! d x = ∫ 0 1 x 2 e x d x = [ x 2 e x ] 0 1 − 2 ∫ 0 1 x e x d x = e − 2 [ e − ( e − 1 ) ] = e − 2 \begin{aligned} A & = \lim_{n \to \infty} { \sum_{r=0}^n {\frac{\left( _r^n \right)}{n^r(r+3)}}} \\ & = \lim_{n \to \infty} { \sum_{r=0}^n {\frac{n!}{r!(n-r)!n^r(r+3)}}} \\ & = \lim_{n \to \infty} { \sum_{r=0}^n {\left[ \color{#3D99F6} {\lim_{n \to \infty} \left( \frac{n!}{n^r(n-r)!}\right)} \left( \frac{1}{r!(r+3)} \right)\right]}} \\ & = \lim_{n \to \infty} { \sum_{r=0}^n {\frac{1}{r!(r+3)}}} \quad \quad \color{#3D99F6} {\text{Note: } \lim_{n \to \infty} \left( \frac{n!}{n^r(n-r)!}\right) = 1} \\ & = \lim_{n \to \infty} { \sum_{r=0}^n {\frac{\color{#3D99F6}{x^{r+3}}}{r!(r+3)}}} \quad \quad \color{#3D99F6}{\text{for } x = 1} \\ & = \lim_{n \to \infty} { \sum_{r=0}^n { \int_0^1 {\frac{x^{r+2}}{r!}dx}}} \\ & = \int_0^1 x^2 \lim_{n \to \infty} \sum_{r=0}^n {\frac{x^r}{r!}} dx \\ & = \int_0^1 {x^2e^x \space dx} \\ & = \left[x^2e^x\right]_0^1 - 2 \int_0^1 {xe^x \space dx} \\ & = e - 2 [e-(e-1)] \\ & = \boxed{e - 2} \end{aligned}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Can you elaborate on the 4th to 5th step? It seems to me that you wanted to apply the idea of switching the order of summation and limits, which has not been justified. Care has to be taken when doing so.

Ah! Made a mistake while I was finding the sum using binomial and got the answer as e e . Nice!

Kartik Sharma - 6 years ago

@Chew-Seong Cheong This looks like a good approach, but I'm a bit confused about what happened from the 3rd to 5th steps. Any clarification would be appreciated. :)

Brian Charlesworth - 6 years ago

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Thanks for asking. I have added some info.

Chew-Seong Cheong - 6 years ago

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Great, thanks for the clarifications. It took me a while to convince myself that we could extract n ! n r ( n − r ) ! \dfrac{n!}{n^{r}(n - r)!} in this way, (since r r is the summation index), but since the limit you state does indeed hold for any non-negative integer r r this extraction makes sense. Nice solution. :)

Brian Charlesworth - 6 years ago

How to disappearing n(n-1)(n-2)...(n-r+1) with n^r ? suppose n=4 r=2, then 4 x 3 = 4^2 (?)

de azalea - 6 years ago

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