# Thread: [SOLVED] Proofs of Limits

1. ## [SOLVED] Proofs of Limits

I have the proof of lim n-> (1+α/n)^n = e^α. And I need to define ln x = integral from 1 to x of 1/t dt for x > 1. I also want to-

i) Prove that f(t) = 1/t is decreasing for 1 < t < 1+h hence:

1/1+h < 1/t < 1 for 1 < t < 1+h; and h / 1+h < ln(1+h) < h for all h > 0

ii) Prove that α/n+α < ln(1+α/n) < α/n for all α > 0 and n ∈ ℕ
by setting h = α/n in (i.)

iii) Prove that e^(nα/n+α) < (1+(α/n))^n < e^α.

iv) Prove that lim n-> nα/n+α = α

v) Show that lim n-> (1+(α/n))^n = e^a. I also need the theorem that is used.

What I have so far is:

Integral of 1 to 1+h of 1/t dt = ln t (from 1 to 1+h) = ln(1+h)

h/h+1 < ln(1+h) < h

(α/n)/(1+(α/n)) < ln(1+(α/n)) < α/n

α/(1+(α/n)) < n ln(1+(α/n)) < α

lim n-> α = α

lim n-> n ln(1+(α/n)) = α

lim n-> (1+(α/n))^n = e^α

2. I had to dig this out of my notes, and consult online sources, so some of the logic behind this may be a little foggy, but here's the general idea for the proof from what I remember.

$p= (1+\frac{a}{n})^n$
$s = \sum_{k=0}^n \frac{a^k}{k!}= e^a$

The above is a definition of e.

Express it using binomial theorem:

$p = \sum_{k=0}^n$ $n\choose{k}$ $\frac{a^k}{n^k}$

But when k =0 the function evaluates to 1 and when k = 1 the function evaluates to a, so:

$p = 1 + x + \sum_{k=2}^n$ $n\choose{k}$ $\frac{n(n-1)(n-2)...n-(k-1))a^k}{k!n^k}$

Now you'll be able to cancel part of the bottom and part of the top, which will give you a slightly clearer way to express the entire thing.

From this point the idea is to prove that p < s. Make sure to take a > 0 at this point. For the left side you have to do the same thing, but with the intent of proving that s < p and since s = $e^a$ you'll have an answer.

The steps for the left side of the equation can be found online, but I don't know how to explain them properly, so if someone could pick up where i left off, that would be nice.