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Show that $$\lim \limits_{n \to \infty}\frac{\big(1^{1^p}2^{2^p}\cdots n^{n^p}\big)^{1/(n^{p+1})}}{n^{1/(p+1)}}=e^{-1/(p+1)^2},p\ge0.$$

A hint was given for $p=2.$ It was to use Lagrange mean value theorem to the function $f(x)=\frac{x^3\ln x}{3}-\frac{x^3}{9}$ on the interval $[k,k+1],1\le k\le n.$
I am not able to relate the hint in anyway to the question. Some help please!

Bijesh K.S
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1 Answers1

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I took a different route. Applying $\ln$ to your expression gives

$$\tag 1 \frac{1}{n^{p+1}}\sum_{k=1}^{n} k^p\ln k - \frac{\ln n}{p+1}.$$

Let $S_n = \sum_{k=1}^{n} k^p\ln k.$ Note that $x^p\ln x$ is increasing on $[1,\infty).$ So by comparing the integral to sums of areas of rectangles in the usual way, we get

$$\int_1^n x^p\ln x\, dx < S_n < \int_1^n x^p\ln x\, dx + (n+1)^p\ln (n+1).$$

Now $\int x^p\ln x \, dx = (x^{p+1}\ln x)/(p+1)-x^{p+1}/(p+1)^2.$ Evaluating the integral above then gives

$$\frac{n^{p+1}\ln n}{p+1} -\frac{n^{p+1}}{(p+1)^2} < S_n < \frac{n^{p+1}\ln n}{p+1} -\frac{n^{p+1}}{(p+1)^2} +(n+1)^p\ln n$$

Look back at $(1)$ to see we should now divide everything by $n^{p+1},$ then subtract $(\ln n)/(p+1).$ We get

$$-\frac{1}{(p+1)^2} < \frac{1}{n^{p+1}}S_n - \frac{\ln n}{p+1} < -\frac{1}{(p+1)^2} + \frac{(n+1)^p}{n^{p+1}}\ln (n+1).$$

The last very last term on the right $\to 0,$ so we've shown the limit of $(1)$ is $-1/(p+1)^2.$ Exponentiating back gives the desired limit of $e^{-1/(p+1)^2}.$

zhw.
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  • I think your route is actually essentially the same. If you extend to the generalization of the function given in the hint, one would be inclined to consider the function $\frac{x^{p+1} \log(x)}{p+1} - \frac{x^{p+1}}{(p+1)^2}$. But of course the derivative of this is $x^p \log x$ (and this pops up in your solution), and the rest is pretty much identical to your proof! – Rellek Nov 17 '16 at 21:31