An approximation for e

[Gene Partlow]

I'm an amateur and I just noodled this oddity:

   Let a+1 = b.  As a,b —> infinity, (a^b) / (b^a) ≈ (a+b)/2e ,

where e=2.718281… .

Or, to restate: If a+1=b, e ≈  (a+b)(b^a) / 2 (a^b) , as a,b —> infinity .

Is this a known result?  Eg:  Does it follow from some known

series results?

[Gene Partlow]

Zelos Malum emailed me, saying, "Before going on, i suggest you remove b and just write a+1 and the obviousness of it will become apparent".

When I do this, I get e ≈  [ (2a+1)(a+1)^a ] / 2a^(a+1) .   Can someone please tell me how this makes the relation obvious?

[delaneyrm]

Using b = a + 1, one can show

ln (a^b / b^a) =  (ln(a / a+1)) / (1/a) + ln(a)

As a -> inf. the first term on the right takes the form 0/0, so using L'Hospital's Rule one can show the first term goes to -1. So

lim(a->inf) ln (a^b / b^a) = -1 + ln(a)

and thus

lim(a->inf) a^b / b^a = lim(a->inf) e^(-1 + ln(a)) = lim(a->inf) a/e

Notice that with b = a + 1, Gene’s result (a+b)/2e becomes

lim(a->inf) (a+b)/2e = lim(a->inf) (2a+1) / 2e = lim(a->inf) (a/e + 1/2e) = lim(a->inf) a/e

So there is agreement.

Bob