# Prove that $\lim_{x\to0} ({1^{(1/\sin^2x)} + 2^{(1/\sin^2x)} + 3^{(1/\sin^2x)} + ....+ n^{(1/\sin^2x)})^{\sin^2(x)}} = \frac{n(n+1)}{2}$

$$\lim_{x\to0} ({1^{(1/\sin^2x)} + 2^{(1/\sin^2x)} + 3^{(1/\sin^2x)} + ....+ n^{(1/\sin^2x)})^{\sin^2(x)}} = \frac{n(n+1)}{2}$$

Attempt:

According to me, it has to be $$1$$, since the outermost exponent tends to $$0$$.

But anyways, would taking the limit as equal to $$L$$. And taking ($$\log$$) on both sides help?

• The use of $\sin^2 x$ in this expression seems strange. You could replace $y = \sin^2 x$ and look at the limit as $y \to 0_+$; I'm pretty sure it would be the same. Commented Dec 8, 2016 at 17:02
• It would still be a $(1^{1/y}+... n^{1/y})^y$ limit. Where can I go from here? Commented Dec 8, 2016 at 17:05
• Also, are you sure that this equation is correct? Plotting the left-hand side as a function of $x$ (or $y$) seems to imply that the limit is $n$, not $n(n+1)/2$. Commented Dec 8, 2016 at 17:08
• Yes. I am positive it is correct since I saw this on a worksheet I found online. Can you tell me how did you plot it? Commented Dec 8, 2016 at 17:15
• Here it is on Wolfram Alpha for n = 4. Here it is for n = 10. These aren't proofs, of course, but they're perhaps indicative that there's a problem with the equation as stated. Commented Dec 8, 2016 at 17:23

Let us consider $n=2$ first, for simplicity, and use $x^2$ instead of $\sin^2x$, as $x\to0$. As you suggested, since the function has the indeterminate form $\infty^0$ as $x\to0$, let us consider the logarithm of the original function $$x^2\log\left(1+2^{1/x^2}\right) = \frac{\log\left(1+2^{1/x^2}\right)}{1/x^2},$$ which, as $x\to 0$, is of the form $\infty/\infty$. We can apply l'Hospital's rule getting $$\frac{\log 2}{1+2^{-1/x^2}},$$ which tends to $\log 2$ as $x\to0$. Therefore $$\lim_{x\to 0} \left(1+2^{1/x^2}\right)^{x^2}=2.$$ In general, we may apply l'Hospital's rule to $$\frac{\log\left(1+2^{1/x^2}+3^{1/x^2}+\ldots+n^{1/x^2}\right)}{1/x^2}$$ getting $$\frac{\left(\frac{2}{n}\right)^{1/x^2}\log 2+\left(\frac{3}{n}\right)^{1/x^2}\log 3+\ldots+\log n}{1+\ldots + \left(\frac{3}{n}\right)^{1/x^2}+\left(\frac{2}{n}\right)^{1/x^2}+n^{-1/x^2}},$$ which tends to $\log n$ as $x\to0$, proiving $$\lim_{x\to0}\left(1+2^{1/x^2}+3^{1/x^2}+\ldots+n^{1/x^2}\right)^{x^2}=n.$$

For any $$n$$ numbers $$a_1, \ldots, a_n > 0$$, we have

\begin{align} \lim_{p\to+\infty} (a_1^p + a_2^p + \ldots + a_n^p)^{1/p} &= \max\{ a_1, \ldots, a_n \}\tag{*1a}\\ \lim_{p\to-\infty} (a_1^p + a_2^p + \ldots + a_n^p)^{1/p} &= \min\{ a_1, \ldots, a_n \}\tag{*1b} \end{align}

I will only justify $$(*1a)$$. Relabeling the numbers if necessary, we only need to study the case $$0 < a_1 \le a_2 \ldots \le a_n$$. For any $$p > 0$$, we have

$$a_n = (a_n^p)^{1/p} \le (a_1^p + \ldots + a_n^p)^{1/p} \le (a_n^p + \ldots + a_n^p)^{1/p} = n^{1/p} a_n$$ Since $$\lim\limits_{p\to+\infty} n^{1/p} = 1$$, by squeezing, $$(*1a)$$ follows.

Apply this to the case $$(a_1,\ldots,a_n) = (1,\ldots,n)$$ and notice as $$x \to 0$$, $$\frac{1}{\sin^2x} \to +\infty$$, we get

$$\lim_{x\to 0} \left(1^{1/\sin^2 x} + \ldots + n^{1/\sin^2 x}\right)^{\sin^2 x} = \max\{ 1, \ldots, n \} = n$$