Please critic my proof (Limit converges to e)

[moo5003]

Homework Statement

Prove that 1 + 1 + 1/2! + 1/3! +1/4! + 1/5!... + 1/n!
Limit of this as n goes to infinity is equal to e.

Homework Equations

I already showed (1 + 1/n)^n = e

The Attempt at a Solution

My proof is as follows:

e = limit (n to infinity) Sum (k=1 to n) of:

(n!/(n-k)!) * 1/(k!*n^n)

= limit (n to infinity) of:

1 + n/n + n(n-1)/n^2 * 1/2! + n(n-1)(n-2)/n^3 * 1/3! + ...

Note: each term has the same leading degree and coefficient, meaning each terms tends to 1 * 1/k! as n approaches infinity thus this entire series tends to:

1 + 1 + 1/2! + 1/3! + 1/4!...

My main question/concern is the fact that I passed the limit through the summation. It this ok? I know that it works for finite sums:

ie: Limit (X_n + Y_n) = Limit(X_n) + Limit(Y_n)

but I'm unsure if this is true for infinite sums. Any help would be appreciated.

 

[Dick]

You mean sum (k=1 to n) of (n!/(n-k)!) * 1/(k!*n^k), right? You are right to be concerned. No, you can't commute the limit and the sum for infinite series. For example, lim n->infinity sum(k=1 to n) k/n is divergent. sum(k=1 to n) lim n->infinity k/n=0. But your argument does mean it's reasonable to think sum(k=0 to infinity) 1/k! is e. Given you've just proven that lim (n->infinity) (1+1/n)^n is e, that really might be all they are asking for.

 

[fmam3]

Suppose you're allowed to use the familiar property that \frac{\mathrm{d} \mathrm{e}^x}{\mathrm{d} x} = \mathrm{e}^x. Then first find the Taylor series of the function \mathrm{e}^x and prove that the Lagrange remainder R_n(x) \to 0 and thereby showing that \mathrm{e}^x = \sum_{k = 0}^{\infty} \frac{1}{k!}x^k for \forall x \in \mathbb{R}. Set x = 1 and that's your result.

 

[Dick]

Suppose you're allowed to use the familiar property that \frac{\mathrm{d} \mathrm{e}^x}{\mathrm{d} x} = \mathrm{e}^x. Then first find the Taylor series of the function \mathrm{e}^x and prove that the Lagrange remainder R_n(x) \to 0 and thereby showing that \mathrm{e}^x = \sum_{k = 0}^{\infty} \frac{1}{k!}x^k for \forall x \in \mathbb{R}. Set x = 1 and that's your result.

Sure, that works. But I'm not sure this is supposed to use taylor series, is it moo5003?

 

[lurflurf]

You are allowed to move that limit though the sum, but first you need to justify it. Once it has been justified you may not need to commute them. A common theme. The usual trick is to break the sum up.
let n and m be large with m<<n
then the first m terms of the sum look like those of Σ1/k! and the others are small