Cfrgtkky

Suppose $a_n$ and $b_n$ are sequences of positive numbers, such that
$$
lim_{ntoinfty}a_n^n = a,quad lim_{ntoinfty}b_n^n = b,qquad a,bin (0, infty).
$$

Find limit
$$
lim_{ntoinfty}(pa_n + qb_n)^n,
$$
where $p, q$ are nonnegative numbers such that $p + q = 1$.

We use the following lemmas:

Lemma 1: Let ${a_n} $ be a sequence such that $a_nto a>0$. Then $n(a_n^{1/n}-1)tolog a$ (prove it).

Lemma 2 : Let ${a_n} $ be a sequence such that $n(a_n-1)to 0$. Then $a_n^nto 1$.

Consider the sequence ${x_n} $ defined by $$x_n=frac{pa_n+qb_n} {a_n^pb_n^q} $$ and then we have $$n(x_n-1)=frac{n((a_n^{np}
b_n^{nq}) ^{1/n}-1)-pn((a_n^n)^{1/n}-1)-qn((b_n^n)^{1/n}-1)}{a_n^pb_n^q}$$ By lemma $1$ numerator of the above expression tends to $$log(a^pb^q) - plog a-qlog b=0$$ and since $a_nto 1,b_nto 1$ (prove this) the denominator tends to $1$ so that $n(x_n-1)to 0$. Thus by lemma 2 the sequence $x_n^nto 1$ and thus the desired limit is equal to $a^pb^q$.

Alternatively we can use lemma 1 and the following

Lemma 3: Let ${a_n} $ be a sequence such that $a_nto a$ then $$left(1+frac{a_n}{n}right)^nto e^a$$

We can write $$pa_n+qb_n=1+frac{x_n}{n}$$ where $$x_n=pn((a_n^{n})^{1/n}-1)+qn((b_n^{n})^{1/n}-1)$$ and by lemma 1 $$x_nto plog a+qlog b=log(a^pb^q) $$ and therefore by lemma 3 we have $$(pa_n+qb_n)^n=left(1+frac{x_n}{n}right)^nto exp(log(a^pb^q)) =a^pb^q$$

Write $a_n^n = e^{alpha_n}$ and $b_n^n = e^{beta_n}$. Then $alpha_n to log a$ , $beta_n to log b$ and

$$ a_n = 1 + frac{alpha_n}{n} + mathcal{O}left(frac{1}{n^2}right), qquad b_n = 1 + frac{beta_n}{n} + mathcal{O}left(frac{1}{n^2}right). $$

Therefore

$$ (p a_n + q b_n)^n = left( 1 + frac{p alpha_n + q beta_n}{n} + mathcal{O}left(frac{1}{n^2}right) right)^n xrightarrow[ntoinfty]{} e^{p log a + q log b} = a^p b^q. $$