Elementary Calculus: Geometric Series Formula

We conclude this section with the proof of the lemma that lim_x→∞ (1 + 1/x)^xexists. We use the following formula from elementary algebra.

(1 + b + b² + ... + bⁿ)(b - 1) = (b + b² + ... + bⁿ + bⁿ⁺¹) - (1 + b + ... + b^n-1 + bⁿ) = bnⁿ⁺¹ - 1.

PROOF OF THE LEMMA

The function y = 2^t is continuous and positive. Therefore the integral

is a positive real number. Our plan is to use the fact that the Riemann sums approach c to show that (1 + 1/x)^x approaches the limit 2^c.

Let H be positive infinite. We wish to prove that

It is easier to work with the logarithm

Let

Δt is positive and is infinitesimal because

Δt ≈ log₂ 1 = 0.

Moreover,

(3)

Let us form the Riemann sum

For simplicity suppose Δt evenly divides 1, so K Δt = 1. By the Geometric Series Formula,

By Equation 3,

Taking standard parts we have

Finally,

The proof is the same when Δt does not evenly divide 1, except that K Δt is infinitely close to 1 instead of equal to 1. Therefore

We remark that in the above proof we could have used any other positive real number in place of 2. Notice that 2^c = e, so the constant

is just log₂ e.

Home Exponential and Logartihmic Functions Derivatives of Exponential Functions and the Number e Geometric Series Formula - Proof of Lemma


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Geometric Series Formula - Proof of Lemma We conclude this section with the proof of the lemma that lim_x→∞ (1 + 1/x)^xexists. We use the following formula from elementary algebra. GEOMETRIC SERIES FORMULA If b ≠ 1, then This formula is proved by multiplying (1 + b + b² + ... + bⁿ)(b - 1) = (b + b² + ... + bⁿ + bⁿ⁺¹) - (1 + b + ... + b^n-1 + bⁿ) = bnⁿ⁺¹ - 1. PROOF OF THE LEMMA The function y = 2^t is continuous and positive. Therefore the integral is a positive real number. Our plan is to use the fact that the Riemann sums approach c to show that (1 + 1/x)^x approaches the limit 2^c. Let H be positive infinite. We wish to prove that It is easier to work with the logarithm Let Δt is positive and is infinitesimal because Δt ≈ log₂ 1 = 0. Moreover, so (3) Let us form the Riemann sum For simplicity suppose Δt evenly divides 1, so K Δt = 1. By the Geometric Series Formula, By Equation 3, Taking standard parts we have Finally, The proof is the same when Δt does not evenly divide 1, except that K Δt is infinitely close to 1 instead of equal to 1. Therefore We remark that in the above proof we could have used any other positive real number in place of 2. Notice that 2^c = e, so the constant is just log₂ e.
Home Exponential and Logartihmic Functions Derivatives of Exponential Functions and the Number e Geometric Series Formula - Proof of Lemma

Geometric Series Formula - Proof of Lemma