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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0Is there anything special about $a$? I'm assuming it is an arbitrary real number? – 2012-02-22
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1Welcome to math.SE! Do you have learned Riemann integration? – 2012-02-22
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4You may prove a more general statement: *If $a_n \to a$, then $\frac{1}{n}\sum_{k=1}^{n}a_k \to a$*. This can be proved by $\epsilon-\delta$ argument. – 2012-02-22
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0a is an arbitrary real number, yeah. And I've learned Riemann integration! – 2012-02-22
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0@sos440: Thanks, I thought I could try something like this. But I'm wondering if there's perhaps a 'special' trick with this particular limit? leo mentions Riemann integration. – 2012-02-22
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0I think that leo jumped the gun a bit: that sum doesn’t actually appear to be a Riemann sum. – 2012-02-22
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0Working in that. I'm dealing with the $k$... – 2012-02-22
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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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0Based on Brian and sos's arguments, I was wondering if I'm correctly generalizing this situation: Say we have $a_n->a$, and let's take some arbitrary $l>0$ (using 'l' instead of epsilon). Then $|(\frac{1}{n}\sum_{k=1}^{n}a_n)-a|=|\frac{1}{n}\sum_{k=1}^{n}(a_n-a)|\leq \frac{1}{n}\sum_{k=1}^{n}|a_n-a|$. For almost every n, we have that $|a_n-a|
, yielding that $\frac{1}{n}\sum_{k=1}^{n}|a_n-a| for almost every n. So $\frac{1}{n}\sum_{k=1}^{n}a_n \rightarrow a$. Does this sound correct? Thanks! – 2012-02-22 -
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 $\displaystyle \frac{1}{n} \sum_{k=1}^{n} k \sin \left( \frac{a}{j} \right) \to a.$