If a_n=(1+\frac{2}{n})^n , then find \lim_{n\to\infty}(1-\frac{a_n}{n})^n

Anika Osborne

Anika Osborne

Answered question

2022-01-24

If an=(1+2n)n , then find
limn(1ann)n

Answer & Explanation

Howard Gallagher

Howard Gallagher

Beginner2022-01-25Added 13 answers

Due to
(1ann)n=[(1ann)nan]ahnn=[(1ann)nan]an
limn(1ann)nan=e
and limn(an)=e2
Let An=(1ann)nan and Bn=an, by the claim below, you can get the result!
sineurosi0f

sineurosi0f

Beginner2022-01-26Added 14 answers

This is a very nice problem; but it does not file under ''composite limits'', i.e., limits of the form xζf(g(x)).
It is agreed that limnan=limn(1+2n)n=e2
The function ulog(1u) is analytic for |u|<1. Therefore ''by Taylor's Theorem'' there is a function ug(u), analytic for |u|<1, such that
log(1u)=u+u2g(u) (1)
It follows that there is an M>0 such that
|g(u)|M
If x and y are arbitrary real numbers with 0<|xy|<1 then we get from (1) that
log(1xy)x=y+xy2g(xy)
Now put x=1n, y=an. Then the last equation says
nlog(1ann)=ab+1nan2g(ann) (2)
whereby this holds as soon as n is large enough to make |an|<n
Now let n. Then ane2 and ann0. Therefore the RHS of (2) converges to e2. So does the LHS, and we get what we have conjectured all along:
limn(1ann)n=exp(e2)

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