`z in CC` , for any `NinNN` we need to show `lim_(n->oo)sum_(k=1)^N [ z^k /(k!) - (n!)/(k!(n-k)!) * z^k / n^k ] = 0`from this we can conclude (1+z/n)^n converges to exp(z) for n->inf

txmedteach | High School Teacher | (Level 3) Associate Educator

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Ha! And here I was feeling bad because it looked like it converged to "not zero."

Alright, that makes a ton more sense. Here's how to prove it.

Our sum is:

`sum_(k=0)^N(z^k/(k!) - ((n),(k))*z^k/n^k)=sum_(k=0)^N(z^k/(k!) - (n!)/(k!(n-k)!)*z^k/n^k)`

Recognize now that we can do some rearranging:

`sum_(k=0)^N(z^k/(k!) - (n!)/(n^k(n-k)!)*z^k/(k!))`

Now, let's consider that if `(n!)/(n^k(n-k)!)` approaches 1 as `n->oo`, then the inside of our sum becomes 0. Then, our whole sum becomes zero.

So, let's show that that limit is 1:

`lim_(n->oo) (n!)/(n^k(n-k)!) = lim_(n->oo) (n(n-1)(n-2)...(n-k+1)(n-k)!)/(n^k(n-k)!)`

Now, we cancel the (n-k)!

`=lim_(n->oo) (n(n-1)(n-2)...(n-k+1))/n^k`

First, notice that there are k terms in the numerator (by virtue of us being left with (n-k)! before cancelling). Now, notice that in each term in the numerator, the "n" term will dominate each expression as `n->oo`, so our limit becomes:

`=lim_(n->oo) (n*n*...*n)/n^k = lim_(n->oo)n^k/n^k = 1`

There you go! we just showed that that big ol' mess in the summation term, as `n->oo` becomes:

`sum_(k=0)^N z^k/(k!) - z^k/(k!)`

If that's not zero, I don't know what is!

Therefore, our overall summary statement that we just proved:

`lim_(n->oo) sum_(k=0)^N z^k/(k!) - (n!)/(k!(n-k)!)*z^k/n^k = 0`

I hope that works out for you!

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