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Is the answer 1 or 2? ($L$ does come out as $\frac 12$.)

My try - Using definite integration as a limit of sum:

$\begin{align*} L &= \lim_{n\to\infty}\Sigma_{r=1}^n \frac{\frac rn \frac 1n}{1+\frac {r}{n^3}} \tag{1} \\&= \int_0^1 xdx \tag{$\frac 1n = dx, \ \frac rn = x$ (2)} \\&= \frac 12 \end{align*}$

Here, my argument: as we are using the right hand values from (1) to (2), and the denominator will be greater than 1 to a much lesser extent than the numerator, $L$ will be slightly greater than $\frac 12$. Therefore, $\frac 1L$ wil be slightly less than 2, and so, the answer should be 1. However, it is given as 2.

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    $\lim_{x\to a}\lfloor x\rfloor$ is different from $\lfloor\lim_{x\to a} x\rfloor$ Whenever $a$ is integer because of the discontinuity.Though since $L<1/2$ the limits are the same if for example you had $L>\frac{1}{2}$ then the limits would be different. – 2017-01-15

2 Answers 2

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Since $\dfrac{rn}{r^3+n^3}\leq\dfrac{rn}{r+n^3}<\dfrac{r}{n^2}$, it follows that $$0.37...=\int_0^1\dfrac{x}{1+x^3}dx\leq L\leq\int_0^1xdx = 0.5$$.

So $\dfrac{1}{L}$ is strictly between $2$ and $3$, and we are done.

EDIT.

If the LHS integral is not simple enough to handle, we can get a looser bound by considering $$\dfrac{rn}{r+n^3} = \dfrac{r}{\tfrac{r}{n}+n^2}\geq\dfrac{r}{r^2+n^2}$$, so the lower bound will be $$\int_0^1\dfrac{x}{1+x^2}dx = \dfrac{\log2}{2} = 0.3465...>\dfrac{1}{3}.$$

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For $n>1$ we have $$\frac {1+n^{-1}}{2(1+n^{-2})}=\frac {n(n^2+n)/2}{n^3+n}=\sum_{r=1}^n\frac {rn}{n^3+n}<$$ $$< \sum_{r=1}^n\frac {rn}{n^3+r}< \sum_{r-1}^n\frac {rn}{n^3}=\frac {n(n^2+n)/2}{n^3}=\frac {1+n^{-1}}{2}.$$ If , instead, you want $\lim_{n\to \infty}$ floor $(1/L_n)$ where $L_n=\sum_{r=1}^nrn/(n^3+r):$ For $n>1$ we have $1/L_n<2(1+n^{-2})/(1+n^{-1})<2.$ And $1/L_n\to 2$ as $n\to \infty.$

So $\lim_{n\to \infty}$ floor $(1/L_n)=1.$

Your justification for the integration is vague. Observe that $$0<\sum_{r=1}^nrn^{-2}-\sum_{r=1}rn^{-2}/(1+rn^{-3})=\sum_{r=1}^nr^2n^{-5}/ (1+rn^{-3})<$$ $$<\sum_{r-1}^nr^2n^{-5}=(n+1)(2n+1)/6n^4$$ and that (obviously) $\lim_{n\to \infty}\frac {1}{n}\sum_{r=1}^nr/n=\int_0^1xdx=1/2.$... And we don't really need the exact value of $\sum_{r=1}^n r^2.$ It is enough to know that it is $o(n^5)$ as $n\to \infty.$ For example $\sum_{r=1}^nr^2<\sum_{r=1}^n\int_r^{r+1}x^2dx=\int_1^{n+1}x^2dx=((n+1)^3-1)/3.$