Evaluate:$$\lim_{x\rightarrow \infty}\frac{\ln \left( \sqrt[x]{a_1^x+a_2^x+...+a_n^x}-a_1 \right)}{x},\ a_1\ge a_2\ge ...\ge a_n>0$$ Lop does not work. I have no idea about this limit.
I can prove $\ln(a_1^x+a_2^x+...+a_n^x)\sim x\ln a_1$ but I got stuck here.

  • $\begingroup$ If you take a look at the top, with large value of $x$, the function inside the root will go towards $a_1$ from the positive side. Subtracting an additional $a_1$ shows that the numerator is approximating $\ln(0)$ from the positive side, or approximately $-\infty$. The bottom approximates $\infty$. Therefore, I believe we can bound the limit as follows: $-1 \le L \le 0$. I can't find anything further as of now. $\endgroup$ – John Lou Apr 18 '18 at 14:55
  • $\begingroup$ I would hazard that the limit goes to $0$, as $x$ goes to infinity far faster than the numerator, I think, although I can't prove it. $\endgroup$ – John Lou Apr 18 '18 at 14:58

The limit is $\ln \frac{a_2}{a_1}$ Changing the notation to something I like better I show

$$\lim\limits_{n\to \infty}\frac{\ln( \sqrt[n]{a_1^n+\cdots +a_k^n}-a_1)}{n}=\ln \frac{a_2}{a_1}$$

Now $$\frac{\ln( \sqrt[n]{a_1^n+\cdots +a_k^n}-a_1)}{n}=\frac{\ln a_1}{n}+\frac{\ln( \sqrt[n]{1+(\frac{a_2}{a_1})^n+\cdots +(\frac{a_k}{a_1})^n}-1)}{n}$$ So it suffices to show $$\lim\limits_{n\to \infty} \frac{\ln( \sqrt[n]{1+(\frac{a_2}{a_1})^n+\cdots +(\frac{a_k}{a_1})^n}-1)}{n} =\ln \frac{a_2}{a_1}$$ Or taking exponents that

$$\lim\limits_{n\to \infty} ( \sqrt[n]{1+(\frac{a_2}{a_1})^n+\cdots +(\frac{a_k}{a_1})^n}-1))^{\frac{1}{n}} = \frac{a_2}{a_1}$$ So now we SQUEEZE. $$( \sqrt[n]{1+(\frac{a_2}{a_1})^n+\cdots +(\frac{a_k}{a_1})^n}-1))^{\frac{1}{n}} \leq ( 1+(\frac{a_2}{a_1})^n+\cdots +(\frac{a_k}{a_1})^n-1))^{\frac{1}{n}}\to \frac{a_2}{a_1} $$

On the other hand letting $p=(\frac{a_2}{a_1})^n+\cdots +(\frac{a_k}{a_1})^n$ $$\frac{p}{n(1+p)}\leq \frac{p}{(\sqrt[n]{1+p})^{n-1}+\cdots +1}=\sqrt[n]{1+p}-1 $$ So

$$\frac{\sqrt[n]{p}}{\sqrt[n]{n}\sqrt[n]{(1+p)}}\leq( \sqrt[n]{1+(\frac{a_2}{a_1})^n+\cdots +(\frac{a_k}{a_1})^n}-1))^{\frac{1}{n}}$$

And$$\frac{\sqrt[n]{p}}{\sqrt[n]{n}\sqrt[n]{(1+p)}}=\frac{\sqrt[n]{(\frac{a_2}{a_1})^n+\cdots +(\frac{a_k}{a_1})^n}}{\sqrt[n]{n}\sqrt[n]{(1+(\frac{a_2}{a_1})^n+\cdots +(\frac{a_k}{a_1})^n)}}\to\frac{a_2}{a_1}$$

| cite | improve this answer | |

We have

$$\ln \left( \sqrt[x]{a_1^x+a_2^x+...+a_n^x}-a_1 \right)=\ln \left( a_1\sqrt[x]{1+(a_2/a_1)^x+...+(a_n/a_1)^x}-a_1 \right)=\\ =\ln a_1+\ln \left( \sqrt[x]{1+(a_2/a_1)^x+...+(a_n/a_1)^x}-1 \right)$$


$$\sqrt[x]{1+(a_2/a_1)^x+...+(a_n/a_1)^x}=e^{\frac{\log(1+(a_2/a1)^x+...+(a_n/a_1)^x)}{x}}\sim e^{\frac{(a_2/a1)^x+...+(a_n/a_1)^x}{x}}\\\sim 1+\frac{(a_2/a_1)^x+...+(a_n/a_1)^x}{x}$$


$$\ln \left( \sqrt[x]{1+(a_2/a_1)^x+...+(a_n/a_1)^x}-1 \right)\sim \ln \left(\frac{(a_2/a_1)^x+...+(a_n/a_1)^x}{x}\right)=\\=\ln \left((a_2/a_1)^x+...+(a_n/a_1)^x\right)-\ln x$$

and finally

$$\frac{\ln \left( \sqrt[x]{a_1^x+a_2^x+...+a_n^x}-a_1 \right)}{x}\sim \frac{\ln a_1+\ln \left((a_2/a_1)^x+...+(a_n/a_1)^x\right)-\ln x}x\\\sim \frac{\ln \left((a_2/a_1)^x+...+(a_n/a_1)^x\right)}x$$

therefore the result depends upon the particular values for the coefficients $a_i$.

| cite | improve this answer | |
  • $\begingroup$ Doesn't the requirement of $a_{1} \ge a_{n} \ge a_{n+1} \ge 0$ help solve the last to $0$? $\endgroup$ – John Lou Apr 18 '18 at 15:06
  • $\begingroup$ @JohnLou I've plugged in the approximation obtained in the line before $\endgroup$ – user Apr 18 '18 at 15:06
  • $\begingroup$ @JohnLou no since it depends upon the actual values for $a_i$ fo example $a_1=a_2=2$ and $a_3=1$ leads to $0$ while $a_1=3, a_2=2$ and $a_3=1$ leads to $\ln 2/3$. $\endgroup$ – user Apr 18 '18 at 15:09
  • 1
    $\begingroup$ @JohnLou sorry there was a big typo in the second line, now it should be fine $\endgroup$ – user Apr 18 '18 at 15:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.