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?

  • $\begingroup$ I think both inequalities should be reversed $\endgroup$ – angryavian Oct 14 '19 at 23:16
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    $\begingroup$ I think you mean $\sqrt{a_n b_n} \le \frac{a_n + b_n}{2} \le \max(a_n, b_n)$. $\endgroup$ – angryavian Oct 14 '19 at 23:18
  • $\begingroup$ Do you mean $a_n>b_n$ or $a_n>0$? $\endgroup$ – kingW3 Oct 14 '19 at 23:42
  • $\begingroup$ I am so sorry, there was a typo $\endgroup$ – user3503589 Oct 14 '19 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 $\lim\limits_{x\to 1}\dfrac{\log x} {x-1}=1$.

  • $\begingroup$ 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? $\endgroup$ – user Oct 15 '19 at 8:32
  • $\begingroup$ @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 $\lim\limits _{x\to 1}\dfrac{\log x} {x-1}=1$. $\endgroup$ – Paramanand Singh Oct 15 '19 at 8:37
  • $\begingroup$ Thanks! I can’t see that, now it’s clear! Bye $\endgroup$ – user Oct 15 '19 at 8:48
  • $\begingroup$ 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? $\endgroup$ – Monolite Oct 15 '19 at 16:09
  • $\begingroup$ @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. $\endgroup$ – Paramanand Singh Oct 15 '19 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

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



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

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

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

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    $\begingroup$ why does $a_n^n\to a \implies a_n^n=a+o\left(\frac1n\right)$ ? $\endgroup$ – Monolite Oct 15 '19 at 15:00
  • $\begingroup$ @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. $\endgroup$ – user Oct 15 '19 at 16:44
  • $\begingroup$ 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! $\endgroup$ – Monolite Oct 15 '19 at 17:04
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    $\begingroup$ @Monolite We have that $$n(a_n-1)=\frac{a_n-1}{\frac1n}=\frac{\sqrt[n]{a_n^n}-1}{\frac1n}=\frac{\left(a+o\left(1\right)\right)^\frac1n-1}{\frac1n}=\frac{a^\frac1n-1}{\frac1n}+o\left(1\right) \to \log a$$ $\endgroup$ – user Oct 15 '19 at 17:16

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