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Suppose $a_n$ and $b_n$ are sequences of positive numbers, such that $$ \lim_{n\to\infty}a_n^n = a,\quad \lim_{n\to\infty}b_n^n = b,\qquad a,b\in (0, \infty). $$
Find limit $$ \lim_{n\to\infty}(pa_n + qb_n)^n, $$ where $p, q$ are nonnegative numbers such that $p + q = 1$.

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  • $\begingroup$ When $a>b$, consider $$(pa_n + qb_n)^n=a^n\left(p+q\cdot \frac{b_n}{a_n}\right)^n.$$ $\endgroup$ – user587192 Nov 13 '18 at 1:26
  • $\begingroup$ Very similar: math.stackexchange.com/q/2995279/587192 $\endgroup$ – user587192 Nov 13 '18 at 1:28
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We use the following lemmas:

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

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

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 $$\log(a^pb^q) - p\log a-q\log b=0$$ and since $a_n\to 1,b_n\to 1$ (prove this) the denominator tends to $1$ so that $n(x_n-1)\to 0$. Thus by lemma 2 the sequence $x_n^n\to 1$ 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 $a_n\to a$ then $$\left(1+\frac{a_n}{n}\right)^n\to e^a$$

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 $$x_n\to p\log a+q\log b=\log(a^pb^q) $$ and therefore by lemma 3 we have $$(pa_n+qb_n)^n=\left(1+\frac{x_n}{n}\right)^n\to \exp(\log(a^pb^q)) =a^pb^q$$

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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

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

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  • $\begingroup$ How did you derive a formula for $a_n$ and $b_n$? Is there a way to do this, without asymptotics? $\endgroup$ – user4201961 Nov 13 '18 at 10:14
  • $\begingroup$ @user4201961: this is nothing but Taylor series for $e^x$. Just note that $a_n=e^{\alpha_n/n} $. $\endgroup$ – Paramanand Singh Nov 13 '18 at 14:50

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