The sequence of functions $f_n:[0,1]\to\mathbb R$ $(n\ge2)$ is given by $f_n=1+x^{n^2-1}+x^{n^2+2n}$. Let $S_n$ denote the area of the figure bounded by the graph of the function $f_n$ and the lines $x=0$, $x=1$, and $y=0$. Compute $$\lim_{n\to\infty}\left(\frac{\sqrt{S_1}+\sqrt{S_2}+\ldots+\sqrt{S_n}}n\right)^n.$$
Problem
Source: 2001 Moldova MO Grade 12 P1
Tags: function
oolite
14.04.2021 11:39
This is ill-posed, as $f_1$ (and therefore $S_1$) is undefined. Should the numerator be $\sqrt{S_2}+\sqrt{S_3}+\ldots+\sqrt{S_{n+1}}$ instead?
$e^{7/12}$
, I think.
augustin_p
06.07.2023 12:49
oolite wrote: This is ill-posed, as $f_1$ (and therefore $S_1$) is undefined. Indeed. The numerator is actually $\sqrt{S_2}+\sqrt{S_3}+\ldots+\sqrt{S_{n}}.$
$$S_n=\int_{0}^{1} 1+x^{n^2-1}+x^{n^2+2n} dx = 1+ \frac{1}{n^2} + \frac{1}{(n+1)^2} = \left(\frac{n^2+n+1}{n(n+1)}\right)^2= \left (1+\frac{1}{n}-\frac{1}{n+1}\right)^2.$$Then $$\lim_{n\to\infty}\left(\frac{\sqrt{S_2}+\sqrt{S_3}+\ldots+\sqrt{S_n}}n\right)^n=\lim_{n\to\infty}\left(1-\frac{n+3}{2n(n+1)}\right)^n=e^{\lim\limits_{n\to\infty}\frac{n+3}{2(n+1)}}=e^{\frac{1}{2}}.$$