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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.

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

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We have

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

and

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

then

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

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 $a_i$.

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  • $\begingroup$ 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 $\endgroup$ – gimusi 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 $\ln 2/3$. $\endgroup$ – gimusi Apr 18 '18 at 15:09
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    $\begingroup$ @JohnLou sorry there was a big typo in the second line, now it should be fine $\endgroup$ – gimusi Apr 18 '18 at 15:32

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