Consider the series $$f_n=\sum_{k=0}^{n-1}\frac{\sin^2(x+\frac{k}{n})}{n}.$$ Since $\lim_{n\rightarrow\infty}f_n=\int_x^{x+1}\sin^2(y)dy$, we know that $f_n$ converges.
And my question is: will the sequence converge uniformly?
I tried to integrate term by term, and the resulting series does converge to the integral of the limiting function. So the series might converge uniformly, but I do not know how to check this directly. It looks too bizarre to check directly, so I am wondering if there is any theorem that might come in handy.

In general, is there a criterion telling us when a Riemann sum converges uniformly?

If something above is wrong, tell me. Thanks in advance.

  • 4
    $\begingroup$ "Since this $f_n$ is the Riemann sum of the function $\sin^2(x)$ in the interval $[x,x+1]$, we know that..." Actually we know that you are mixing much too much functions and their values! You might want to explain what you mean by the quoted sentence. $\endgroup$
    – Did
    Jul 26, 2013 at 16:00
  • $\begingroup$ @Did I meant to say that $\lim_{n\rightarrow\infty}f_n=\int_x^{x+1}\sin^2(y)dy$ so that the sequence $f_n$ does converge. $\endgroup$
    – awllower
    Jul 27, 2013 at 1:51

2 Answers 2


Here, the key is going to be the $t\mapsto\sin^2(t)$ is uniformly continuous.

Suppose we're given $\epsilon>0$. Then there exists $\delta>0$ so that $\lvert \sin^2(t)-\sin^2(a)\rvert<\epsilon$ whenever $0<\lvert t-a\rvert<\delta$.

Choose $N\in\mathbb{N}$ large enough that $0<\frac{1}{N}<\delta$. Use uniform continuity to show that for $n\geq N$, for each $0\leq k\leq n-1$, $$ \left\lvert \frac{\sin^2\left(x+\frac{k}{n}\right)}{n}-\int_{x+\frac{k}{n}}^{x+\frac{k+1}{n}}\sin^2(t)\,dt\right\rvert<\frac{\epsilon}{n}, $$ no matter what $x$ you choose. It follows by the triangle inequality that for any $x$, for $n\geq N$ you have $$ \left\lvert\sum_{k=0}^{n-1}\frac{\sin^2(x+\frac{k}{n})}{n}-\int_{x}^{x+1}\sin^2(t)\,dt\right\rvert<\epsilon. $$ In general, when you're dealing with Riemann sums, this will be a good place to start; at the least it will help you think about where possible problems could occur.

  • $\begingroup$ Hence, the Riemann sums of a uniformly continuous functions converges uniformly to the integral, right? Thanks for the answer. :) $\endgroup$
    – awllower
    Jul 27, 2013 at 1:54
  • 1
    $\begingroup$ Yes, that is the case - in fact, you could set up a 2-variable function which would converge uniformly to the integral on $[x,y]$, and use a similar proof that it converged uniformly. $\endgroup$ Jul 27, 2013 at 12:47
  • $\begingroup$ And: my pleasure! $\endgroup$ Jul 27, 2013 at 12:48

A general argument is the following theorem of Dini : An increasing sequence of continuous functions on a compact space such that its limit is still a continuous function, converges not only pointwise but uniformly.

  • $\begingroup$ Interesting to hear that theorem! That is another direction than the above. Thanks for the answer. $\endgroup$
    – awllower
    Oct 21, 2013 at 16:19

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