Cfrgtkky

The title says it all, thanks in advance. I just cannot see how to solve this using binomial theorem, maybe use it twice?

Let $a, b in mathbb{C}$. We begin with the identity

$$ frac{left( 1 + frac{a}{n} + frac{b}{n^2} right)^n}{left(1 + frac{a}{n} right)^n}
= left( 1 + frac{b}{n(n + a)} right)^n. $$

Now let $n > |a|+|b| $. Then by the binomial theorem, together with the simple estimate $binom{n}{k} leq n^k$,

begin{align*}
left| frac{left( 1 + frac{a}{n} + frac{b}{n^2} right)^n}{left(1 + frac{a}{n} right)^n} - 1 right|
&= left| sum_{k=1}^{n} binom{n}{k} frac{b^k}{n^k(n + a)^k} right| \
&leq sum_{k=1}^{n} binom{n}{k} frac{|b|^k}{n^k(n - |a|)^k} \
&leq sum_{k=1}^{n} frac{|b|^k}{(n - |a|)^k} \
&leq frac{|b|}{n - |a| - |b|},
end{align*}

which converges to $0$ as $ntoinfty$. Therefore

$$ lim_{ntoinfty} frac{left( 1 + frac{a}{n} + frac{b}{n^2} right)^n}{left(1 + frac{a}{n} right)^n} = 1, $$

and the desired conclusion follows by assuming that we know the limit $lim left(1 + frac{a}{n} right)^n$ exists in $mathbb{C}setminus{0}$.

If you are assuming that the limits exists then you can argue as follows: $(1+frac a n +frac b {n^{2}})^{n}-(1+frac a n )^{n}= sum_{j=0}^{n-1} binom {n} {j}(1+frac a n )^{l}(frac b {n^{2}})^{n-j}$. Note that $|(frac b {n^{2}})^{n-j}| leq frac {|b|} {n^{2}}$ as long as $frac {|b|} {n^{2}} <1$. The remaining sum is bounded by $(2+frac {|a|} n)^{n}$ which is bounded.

Let us first assume that $a,bge 0$. (The remaining cases $a<0$ can be treated analogously.)

Using the fact that
$$
s=lim_{ntoinfty}left(1+frac{a}{n}right)^nin (0,infty),
$$
we obtain that
$$
1le frac{left(1+frac{a}{n}+frac{b}{n^2}right)^n}{left(1+frac{a}{n}right)^n}leleft(1+frac{b}{n^2}right)^n= left(Big(1+frac{b}{n^2}Big)^{n^2}right)^{1/n}to 1
$$
and hence
$$
lim_{ntoinfty}frac{left(1+frac{a}{n}+frac{b}{n^2}right)^n}{left(1+frac{a}{n}right)^n}=1.
$$

Let $n^2+an+b=(n+p)(n+q)implies p+q=a, pq=b$

$$lim_{ntoinfty}left(1+dfrac{an+b}{n^2}right)^n=lim_{ntoinfty}left(1+dfrac pnright)^nlim_{ntoinfty}left(1+dfrac qnright)^n=e^{p+q}=e^a$$

Now $left(1+dfrac anright)^n=?$