# Compute $\lim_{n \to \infty}(\frac{a_n+b_n}{2})^n$

I am trying to solve the following problem:

Compute $$\lim_{n \to \infty}(\frac{a_n+b_n}{2})^n$$ when $$\lim_{n \to \infty} a_n^n=a>0$$ and $$\lim_{n \to \infty} b_n^n=b>0$$ such that $$a_n,b_n>0 \ \forall \ n \ \in \mathbb{N}$$.

I tried to use the Sandwich Theorem to come up with an answer, but my upper bound was not tight:

$$\max(a_n,b_n)\ge(\frac{a_n+b_n}{2}) \ge \sqrt{a_nb_n}$$

On passing to the limits I got the following:

$$\max(a,b)\ge \lim_{n \to \infty}(\frac{a_n+b_n}{2}) \ge \sqrt{ab}$$

But this doesn't help me at all. How could I actually compute the limit?

• I think both inequalities should be reversed – angryavian Oct 14 at 23:16
• I think you mean $\sqrt{a_n b_n} \le \frac{a_n + b_n}{2} \le \max(a_n, b_n)$. – angryavian Oct 14 at 23:18
• Do you mean $a_n>b_n$ or $a_n>0$? – kingW3 Oct 14 at 23:42
• I am so sorry, there was a typo – user3503589 Oct 14 at 23:44

This is done in steps. First we have to note that both $$a_n, b_n$$ tend to $$1$$. This follows from the fact that $$n\log a_n\to \log a$$ and thus $$\log a_n\to 0$$.

Next we can let $$x_n$$ denote the expression whose limit is to be evaluated here. Then we have $$\log x_n=n\log \left(1+\frac{a_n+b_n-2}{2}\right)$$ and the limit of above expression is same as that of $$\frac{1}{2}\cdot\{n(a_n-1)+n(b_n-1)\}$$ Next we can use the fact that $$n\log a_n\to\log a$$ which implies $$n(a_n-1)\to\log a$$. The limit of $$\log x_n$$ is thus equal to $$\frac{\log a +\log b} {2}$$ It follows that $$x_n\to\sqrt{ab}$$.

The above argument makes use of the standard limit $$\lim\limits_{x\to 1}\dfrac{\log x} {x-1}=1$$.

• How do you suggest to prove that $n(a_n-1)\to \log a$? I’ve used little o notation. Are you thinking to an alternative approach? – user Oct 15 at 8:32
• @user: just note that $$n\log a_n=\frac{\log a_n} {a_n-1}\cdot n(a_n-1)$$ The left hand side tends to $\log a$ and first factor on right tends to $1$ (because $a_n\to 1$) therefore the second factor $n(a_n-1)$ also tends to $\log a$. This is a typical use of limit laws combined with standard limit $\lim\limits _{x\to 1}\dfrac{\log x} {x-1}=1$. – Paramanand Singh Oct 15 at 8:37
• Thanks! I can’t see that, now it’s clear! Bye – user Oct 15 at 8:48
• why can we say that $n \log( 1+ (a_n + b_n -2)/ 2 )$ has the same limit of $\frac{1}{2} ( n (a_n -1) + n (b_n - 1))$? I understand we taylor expand log but why can we disregard the other terms of the taylor series? – Monolite Oct 15 at 16:09
• @Monolite: this is again a typical use of standard limits and limit laws. Write $$n\log(1+x)=nx\cdot\frac{\log(1+x)}{x}$$ where $x=(a_n+b_n-2)/2$ so that $x\to 0$ and thus the fraction $(\log(1+x))/x\to 1$. It follows that $n\log(1+x)$ and $nx$ have same limit. – Paramanand Singh Oct 15 at 16:39

The key point is that

• $$a_n^n\to a \implies a_n^n=a+o\left(1\right)\implies a_n=\sqrt[n]a+o\left(\frac1{n}\right)\to 1$$
• $$b_n^n\to b\implies b_n^n=b+o\left(1\right)\implies b_n=\sqrt[n]b+o\left(\frac1{n}\right)\to 1$$

therefore we have that

$$\left(\frac{a_n+b_n}{2}\right)^n =\left(1+\frac{a_n-1+b_n-1}{2}\right)^n=$$

$$=e^{n\log\left(1+\frac{a_n-1+b_n-1}{2}\right)}\to\sqrt{ab}$$

indeed

$$n\log\left(1+\frac{a_n-1+b_n-1}{2}\right)= \frac12\frac{a_n-1+b_n-1}{\frac1n}\frac{\log\left(1+\frac{a_n-1+b_n-1}{2}\right)}{\frac{a_n-1+b_n-1}{2}}\to \log\sqrt{ab}$$

since by standard limit $$x\to 0,\quad \frac{\log(1+x)}{x}\to 1$$

$$\frac{\log\left(1+\frac{a_n-1+b_n-1}{2}\right)}{\frac{a_n-1+b_n-1}{2}}\to 1$$

and by standard limit $$x\to 0,\quad \frac{A^x-1}{x}\to \log A$$

• $$\frac{a_n-1}{\frac1n}=\frac{\sqrt[n]{a_n^n}-1}{\frac1n}=\frac{\left(a+o\left(1\right)\right)^\frac1n-1}{\frac1n}=\frac{a^\frac1n-1}{\frac1n}+o\left(1\right) \to \log a$$
• $$\frac{b_n-1}{\frac1n}=\dots=\frac{b^\frac1n-1}{\frac1n}+o\left(1\right) \to \log b$$

we have

$$\frac12\frac{a_n-1+b_n-1}{\frac1n}=\frac12\left(\frac{a_n-1}{\frac1n}+\frac{b_n-1}{\frac1n}\right)\to \frac12(\log a + \log b)=\log\sqrt{ab}$$

• why does $a_n^n\to a \implies a_n^n=a+o\left(\frac1n\right)$ ? – Monolite Oct 15 at 15:00
• @Monolite That's a good point! I want indicate a quantity which goes to zero as $n\to \infty$. Thinking again about it, it should be indicated as $o(1)$. I update that. – user Oct 15 at 16:44
• No prob! also, if I may, can I ask you how you proved that $n(a_n -1) \rightarrow \log(a)$ with little o notation? Thanks a lot! – Monolite Oct 15 at 17:04
• @Monolite We have that $$n(a_n-1)=\frac{a_n-1}{\frac1n}=\frac{\sqrt[n]{a_n^n}-1}{\frac1n}=\frac{\left(a+o\left(1\right)\right)^\frac1n-1}{\frac1n}=\frac{a^\frac1n-1}{\frac1n}+o\left(1\right) \to \log a$$ – user Oct 15 at 17:16