I'm trying to calculate $\lim_{n\rightarrow \infty} \frac{1}{n}\sum_{k=1}^{n}k\sin\left(\frac{a}{k}\right)$. Intuitively the answer is $a$, but I can't see any way to show this. Can anyone help? Thanks!
Calculate $\lim_{n\rightarrow \infty} \frac{1}{n}\sum_{k=1}^{n}k\sin\left(\frac{a}{k}\right)$
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0@BrianM.Scott, indeed :-) – 2012-02-22
2 Answers
I’m not going to work out all of the details; rather, I’ll suggest in some detail a way to approach the problem.
First, it suffices to prove the result for $a>0$, since the sine is an odd function. For $a>0$ we have $k\sin\left(\frac{a}k\right)
You know that $\lim\limits_{x\to 0}\frac{\sin x}x=1$, so there is a $c>0$ such that $\sin x>\frac{x}2$ whenever $0
But what I did with $\frac12$ can clearly be done with any positive fraction less than $1$: if $0<\epsilon<1$, there is a $c>0$ such that $\sin x>\epsilon x$ whenever $0
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0@ro44: Not quite. You want $|(\frac1n\sum_{k=1}^na_k)-a|=|\frac1n\sum_{k=1}^n(a_k-a)|\le\frac1n\sum_{k=1}^n|a_k-a|$. Then you want to say that for sufficiently large $n$, ‘most’ terms of this sum are less than $l$, so the sum is less than (say) $2l$. There’s still a bit of work to be done show that if $n$ is large enough, the early ‘bad’ terms are a small enough fraction of the total not to mess things up. – 2012-02-22
$\displaystyle \sin x \leq x $ for $x \geq 0.$ Integrating this over $[0,t]$ gives $ -\cos t +1 \leq \frac{t^2}{2} . $
Integrating both sides again from $[0,x]$ gives $ -\sin x + x \leq \frac{x^3}{6} .$
Thus, $ x - \frac{x^3}{6} \leq \sin x \leq x.$
Hence, $ \frac{1}{n} \sum_{k=1}^{n} k \left( \frac{a}{k} - \frac{a^3}{6k^3} \right ) \leq \frac{1}{n} \sum_{k=1}^{n} k \sin \left( \frac{a}{j} \right) \leq \frac{1}{n} \sum_{k=1}^{n} k \left( \frac{a}{k} \right). $
Since $ \displaystyle \sum_{k=1}^n \frac{a^3}{6k^3} $ is convergent, the Squeeze theorem shows that $\displaystyle \frac{1}{n} \sum_{k=1}^{n} k \sin \left( \frac{a}{j} \right) \to a.$