# How to find $\lim_{n\to\infty}(pa_n + qb_n)^n$ with $p + q = 1$?

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 $$p + q = 1$$.

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

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 $$(pa_n+qb_n)^n=\left(1+\frac{x_n}{n}\right)^n\to \exp(\log(a^pb^q)) =a^pb^q$$

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

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

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