A problem proposed by prof. Babev
Consider the sequence
$$a_n=\int_{n}^{n+1}\frac{\sin^2(\pi t)}{t}\text{d}t.$$
It could be proven that
$$\lim_{n\to \infty}n a_n=\frac{1}{2}$$
Now we are going to prove that
$$\lim_{n\to\infty}n\left(na_n-\frac{1}{2}\right)=-\frac{1}{4} $$
and show how to derive all such limits.
$\textbf{Proof.}$ Since $\displaystyle\int_{n}^{n+1}\sin^2(\pi t)\text{d}{t}=\frac{1}{2}$ (for any $n\in \mathbb{N}$) we can rewrite the limit in question as
$$\lim_{n\to \infty}n\left(n\left(\int_{n}^{n+1}\frac{\sin^2(\pi t)}{t}\text{d}t-\int_{n}^{n+1}\frac{\sin^2(\pi t)}{n}\text{d}t\right)\right)=\lim_{n\to \infty}n^2\left(\int_{n}^{n+1}\sin^2(\pi t)\left(\frac{1}{t}-\frac{1}{n}\right)\text{d}t\right)$$
$$=\lim_{n\to \infty}n\left(\int_{n}^{n+1}\sin^2(\pi t)\frac{n-t}{t}\text{d}t\right)$$
Changing the variables $t\to n+s$ and using periodicity of sine, the last simplifies to
$$ \lim_{n\to \infty}n\int_{0}^{1}\sin^2(\pi s)\frac{-s}{n+s}\text{d}s$$
Since for $s\in[0,1]$, $\displaystyle \frac{n}{n+s}$ is bounded between $1$ and $ \displaystyle \frac{n}{n+1}$, the last limit is bounded between the limits
$$ \lim_{n\to \infty}\int_{0}^{1}\sin^2(\pi s)(-s)\text{d}s\ \ \mbox{and}\ \ \lim_{n\to \infty}\frac{n}{n+1}\int_{0}^{1}\sin^2(\pi s)(-s)\text{d}s$$
which both are equal to $\displaystyle \int_{0}^{1}\sin^2(\pi s)(-s)\text{d}s=-\frac{1}{4}$, hence the result follows.
$\textbf{Comment.}$ In a similar vein it could be proven that
$$\lim_{n\to\infty}n\left(n\left(na_n-\frac{1}{2}\right)+\frac{1}{4}\right)=\int_0^1\sin^2(\pi s)s^2\text{d}s$$
and we could extend this further.
This could be viewed as a "Taylor series of the sequence at infinity", namely
$$a_n=\sum_{j=1}^\infty b_j\left(\frac{1}{n}\right)^j$$
where $$b_j= \int_{0}^{1}\sin^2(\pi s)(-s)^{j-1}\text{d}s$$
Having this perspective leads to general and more insightful solution. Consider the series
$$\sum_{j=1}^\infty b^j \left(\frac{1}{n}\right)^j=\sum_{j=1}^\infty\int_{0}^{1}\sin^2(\pi s)(-s)^{j-1}\left(\frac{1}{n}\right)^j\text{d}s=\int_{0}^{1}\sin^2(\pi s)\frac{1}{n+s}\text{d}s.$$
(where we have exchanged summation and integration and then calculated the sum of the geometric progression)
After changing the variables $s\to t-n$ we arrive at the integral
$$\int_{n}^{n+1}\frac{\sin^2(\pi t)}{t}\text{d}t$$
which is exactly $a_n$.
$$a_n=\int_{n}^{n+1}\frac{\sin^2(\pi t)}{t}\text{d}t.$$
It could be proven that
$$\lim_{n\to \infty}n a_n=\frac{1}{2}$$
Now we are going to prove that
$$\lim_{n\to\infty}n\left(na_n-\frac{1}{2}\right)=-\frac{1}{4} $$
and show how to derive all such limits.
$\textbf{Proof.}$ Since $\displaystyle\int_{n}^{n+1}\sin^2(\pi t)\text{d}{t}=\frac{1}{2}$ (for any $n\in \mathbb{N}$) we can rewrite the limit in question as
$$\lim_{n\to \infty}n\left(n\left(\int_{n}^{n+1}\frac{\sin^2(\pi t)}{t}\text{d}t-\int_{n}^{n+1}\frac{\sin^2(\pi t)}{n}\text{d}t\right)\right)=\lim_{n\to \infty}n^2\left(\int_{n}^{n+1}\sin^2(\pi t)\left(\frac{1}{t}-\frac{1}{n}\right)\text{d}t\right)$$
$$=\lim_{n\to \infty}n\left(\int_{n}^{n+1}\sin^2(\pi t)\frac{n-t}{t}\text{d}t\right)$$
Changing the variables $t\to n+s$ and using periodicity of sine, the last simplifies to
$$ \lim_{n\to \infty}n\int_{0}^{1}\sin^2(\pi s)\frac{-s}{n+s}\text{d}s$$
Since for $s\in[0,1]$, $\displaystyle \frac{n}{n+s}$ is bounded between $1$ and $ \displaystyle \frac{n}{n+1}$, the last limit is bounded between the limits
$$ \lim_{n\to \infty}\int_{0}^{1}\sin^2(\pi s)(-s)\text{d}s\ \ \mbox{and}\ \ \lim_{n\to \infty}\frac{n}{n+1}\int_{0}^{1}\sin^2(\pi s)(-s)\text{d}s$$
which both are equal to $\displaystyle \int_{0}^{1}\sin^2(\pi s)(-s)\text{d}s=-\frac{1}{4}$, hence the result follows.
$\textbf{Comment.}$ In a similar vein it could be proven that
$$\lim_{n\to\infty}n\left(n\left(na_n-\frac{1}{2}\right)+\frac{1}{4}\right)=\int_0^1\sin^2(\pi s)s^2\text{d}s$$
and we could extend this further.
This could be viewed as a "Taylor series of the sequence at infinity", namely
$$a_n=\sum_{j=1}^\infty b_j\left(\frac{1}{n}\right)^j$$
where $$b_j= \int_{0}^{1}\sin^2(\pi s)(-s)^{j-1}\text{d}s$$
Having this perspective leads to general and more insightful solution. Consider the series
$$\sum_{j=1}^\infty b^j \left(\frac{1}{n}\right)^j=\sum_{j=1}^\infty\int_{0}^{1}\sin^2(\pi s)(-s)^{j-1}\left(\frac{1}{n}\right)^j\text{d}s=\int_{0}^{1}\sin^2(\pi s)\frac{1}{n+s}\text{d}s.$$
(where we have exchanged summation and integration and then calculated the sum of the geometric progression)
After changing the variables $s\to t-n$ we arrive at the integral
$$\int_{n}^{n+1}\frac{\sin^2(\pi t)}{t}\text{d}t$$
which is exactly $a_n$.
Задачата, за първите два коефициента, предложих и беше приета за група А на НСОМ 1999 (ИУ Варна). В руски сборник (стандартен за техните ВУЗ) има задача за оценка на интеграла с дробно - линейни функции на n.
ReplyDelete