Proof using the binomial theorem

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SUMMARY

The discussion focuses on proving the equation (1 + 1/n)^n = 1 + sum(k=1 to n) [1/k! product(r=0 to k-1) (1 - r/n)] using the binomial theorem. Participants analyze the transformation of the left-hand side into a sum involving binomial coefficients and factorials. The key steps include rewriting the product on the right-hand side and expressing it in terms of factorials. The conversation highlights the complexity of the proof and the need for clarity in the manipulation of terms.

PREREQUISITES
  • Understanding of the binomial theorem and its applications
  • Familiarity with factorial notation and binomial coefficients
  • Basic knowledge of limits and convergence in calculus
  • Ability to manipulate summations and products in mathematical expressions
NEXT STEPS
  • Study the binomial theorem in-depth, focusing on its proofs and applications
  • Learn about the properties of factorials and their role in combinatorial mathematics
  • Explore convergence of series and limits, particularly in relation to exponential functions
  • Practice manipulating complex mathematical expressions involving sums and products
USEFUL FOR

Students studying advanced mathematics, particularly those focusing on combinatorics and calculus, as well as educators looking for examples of binomial theorem applications.

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Homework Statement


Use the binomial theorem to rpove that for n a positive integer we have:

(1 + 1/n)^n = 1 + sum(k=1 to n) [1/k! product(r=0 to k-1) (1 - r/n)]

The Attempt at a Solution



(1 + 1/n)^n = 1 + sum(k=1 to n) (n choose r) 1^n-k (1/n)^k, where (n choose r) = n!/r!(n - r)!, the binomial coefficients.

I'm trying to fit "n!/r!(n - r)!" to an expression that involves the products, since product(k=1 to n) n = n!

The product on the RHS I rewrite it as: product(r=0 to k-1) [(n - r)/n]

[product(r=0 to k-1) (n - r)] x [product(r=0 to k-1) (1/n)]

=> (n - r)! (1/n)!

So...

(1 + 1/n)^n = 1 + sum(k=1 to n) [1/k! (n-r)! (1/n)!] or (1 + 1/n)^n = 1 + sum(k=1 to n) [(n-r)!/k!n!]

I don't like this. I feel that all of this is taking me nowhere.

Any ideas will be very appreciated.
 
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Hi

look at the snapshot i have attached. in the second equation, its clear that we can express
it as the term on RHS in your expression.
 

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Each of the terms of the RHS sum corresponds to one of the expanded terms of the LHS.
 

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