# Evaluate $\lim\limits_{n\rightarrow \infty}\left\{\frac{1^3}{n^4}+\frac{2^3}{n^4}+\frac{3^3}{n^4}+\dots +\frac{n^3}{n^4}\right\}$

1. Evaluate $\displaystyle\lim_{n\rightarrow \infty}\left\{\frac{1^3}{n^4}+\frac{2^3}{n^4}+\frac{3^3}{n^4}+\dots +\frac{n^3}{n^4}\right\}$.

2. Examine whether $x^{1/x}$ possesses a maximum or minimum and determine the same.

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So far I'm the only person who's up-voted this question. – Michael Hardy Mar 22 '13 at 23:33
@MichaelHardy: It's not exactly a well-posed question. In fact, one might argue that this is not a question at all... – tomasz Mar 23 '13 at 2:25
@tomasz: One might, but it would be casuistry: we all know what questions are being asked here. – Brian M. Scott Mar 23 '13 at 4:10

This begs to be viewed as a Riemann sum: $$\frac 1n\left(\frac{1^3}{n^3}+\cdots+\frac{n^3}{n^3}\right) \to \int_0^1 x^3\,dx.$$

Your second question seems quite different from your first and should be posted separately.

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Ya beat me to it! – Ron Gordon Mar 22 '13 at 18:09

It is happens that there is an attractive closed form for $1^3+2^3+\cdots+n^3$, which can readily be proved by induction: $$1^3+2^3+3^3+\cdots +n^3=\left(\frac{n(n+1)}{2}\right)^2.$$ Divide by $n^4$. Fairly quickly we find that the desired limit is $\dfrac{1}{4}$.

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$${\frac{1^3}{n^4}+\frac{2^3}{n^4}+\frac{3^3}{n^4}+\dots +\frac{n^3}{n^4}}=\frac{\sum_{i=1}^{n}i^3}{n^4}=\frac{(n(n+1))^2}{4n^4}=\frac{1}{4}(1+\frac{1}{n})^2$$

$$\Rightarrow \displaystyle \lim_{n\to\infty}{\frac{1^3}{n^4}+\frac{2^3}{n^4}+\frac{3^3}{n^4}+\dots +\frac{n^3}{n^4}}=\lim_{n\to\infty}\frac{1}{4}(1+\frac{1}{n})^2=1/4$$(Using the fact that $\lim_{n\to\infty}1/n=0)$

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$\displaystyle\lim_{n\rightarrow \infty}\left\{\frac{1^3}{n^4}+\frac{2^3}{n^4}+\frac{3^3}{n^4}+\dots +\frac{n^3}{n^4}\right\}$
= $\displaystyle\lim_{n\rightarrow \infty}\left\{\frac{1^3+2^3+3^3+\dots +n^3}{n^4}\right\}$ = $\displaystyle\lim_{n\rightarrow \infty}\left\{\left(\frac{n(n+1)}{2n^2}\right)^2\right\}$
=$\frac{1}{4}\displaystyle\lim_{n\rightarrow \infty}\left(1+\frac{1}{n^2} \right)^2$
=$\frac{1}{4}$

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Try it another way. – Md Kutubuddin Sardar Mar 22 '13 at 18:20
This answer is very close to and 6 minutes after Abhra Abir Kundu's, so you might not get many votes. Adding some extra value (graphics or extra detail) to your answer might help. – robjohn Mar 22 '13 at 20:32

Hints:

1) This is a Riemann sum.

2) Differentiate and solve for $x$ after equating the first derivative to zero, then try second derivative test to conclude whether the values of $x$ thus obtained is maxima or minima.

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What you need to solve for is $x$. And if you show there's a point where the derivative is $0$, that stops short of showing it's a minimum. – Michael Hardy Mar 22 '13 at 18:10
Yes, but this is just a hint, I didn't want to say everything. – Ishan Banerjee Mar 22 '13 at 18:14