Compute the following limit ($n\in \mathbb{N}$) $$\lim_{x\to \infty}\left(\frac{1}{n}\sum_{k=1}^{n} k^{1/x}\right)^{nx}$$  My idea was to use the inequality:  $$\left(\frac{1}{n}\sum_{k=1}^{n} 1^{1/x}\right)^{nx}&lt;\left(\frac{1}{n}\sum_{k=1}^{n} k^{1/x}\right)^{nx}&lt;\left(\frac{1}{n}\sum_{k=1}^{n} n^{1/x}\right)^{nx} \\ \implies1&lt;L&lt;n^n$$  This gives that the required limit $L$ lies between $1$ and $n^n$. But how can we find its value?
This equation is directly given in my book and I am don't know anything about its proof.I tried L'Hospital rule by differentiating the both numerator as well as denominator(division rule), but the result is still coming in indeterminate forms.I am a beginner , and haven't practiced limits that much. This formula is really confusing me.
I have 3 old computers lying around right now running a P2 at 800 MHz(?), Intel Mobile 1.6 GHz, AMD Athlon XP 2000+ at 1.66 GHz, and (might not use this) P4 at 2.7 GHz, all with 512 MB Ram, and am considering clustering them together for fun/knowledge. They would be running an undecided version of linux, preferably ubuntu based.   The issue is what I want to use it for: general computing and occasional video encoding. By general computing I mean day to day tasks. However I'm not sure if every program started by a single X session is going to exist on the same machine, defeating the purpose of such a system. Will programs be split up or exist on one machine?  Second, assuming this is running 100baseT ethernet (not sure if the PCI slot itself could handle Gigabit), would the speed of having a program exist over the network be an issue? It seems that the constant asking of various things in RAM would be quite slow.  And before you say "buy another computer!", that's not the point of this question. I'm asking would it be usable, not necessarily practical. And yes I know, this is going to be extreamly power consuming. putting a tikzpicture inside a \caption{} I'm attempting to put a small tikzpicture inside a \caption{} belonging to a figure environment. The reason for doing this is fairly simple - to create an in-caption legend for a PGFplot, with consistency between the TikZ styles in the figure and in the caption. Previously I would have done this using dashes and the xcolor package, however, this is an awful hack. In this MWE:  \documentclass{article} \usepackage{tikz} \begin{document} \begin{figure} \caption{\raisebox{0.5ex}{ \begin{tikzpicture}\draw[dashed,color=red,thick] (0,0) -- (0.5,0);\end{tikzpicture}} Gas--Phase}  \end{figure} \end{document}   The expected output:    is generated, but I get ! Argument of \@caption has an extra }. and ! Paragraph ended before \@caption was complete. 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In most introductory quantum mechanics and basic quantum field theory textbooks this repulsive momementum transfer picture is almost always used as a (in classical terms) billiard ball type illustration for electrical repulsion, the Compton effect and so on.  As a photon is a photon is a photon..... I would like to know how it also has the ability to produce atttraction and how negative momentum transfer (such as that between a positively charged object, say a positron and an electron) is explained in any sort of visualisable picture. I'm guessing this may involve Feymann diagrams and the use of time reversal but I'm on shaky ground from there on. I was speaking to an English learner and said, “All you have to do is read a lot.” And they thought that sentence wasn’t grammatically correct because I dropped the word to between is and read.  They thought it should be “All you have to do is to read a lot.” That sounds weird to me.   How do I explain to them the reason you drop the to? Or am I incorrect and it really should have a to? Convergence in Distribution in Order Statistics Let $X_1, X_2, \ldots$ be iid from Exp$(\theta)$ with density function $f(x) = \frac{1}{\theta}e^{-\frac{x}{\theta}}$. (a) Find the limiting distribution of $M_n = Y_1 - \theta\ln(n)$ and $T_n = nY_n$, where $Y_1$ and $Y_n$ represent the first and last order statistics (among $n$ observations) respectively. Answer: I have obtained the CDF of X to be: $$F_x(x) = 1 - e^{-\frac{x}{\theta}}$$ the CDF of $Y_1$ to be: $$F_{Y_1} = 1 = e^{-\frac{xn}{\theta}}$$ the CDF of $M_n$ to be: $$F_{M_n}(x) = \left[1 - \frac{1}{n}e^{-\frac{x}{\theta}}\right]^n$$ with limiting distribution $$e^{-e^{-\frac{x}{\theta}}}$$ the CDF of $T_n$ to be: $$F_{T_n}(x) = 1 - e^{-\frac{x}{\theta}}$$ which is also the limiting distribution. (b) Show that $M_n$ and $T_n$ are asymptotically independent. 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