So this is part of a different problem. The book and my professor say that the series $f(x)=\sum\limits_{k=1}^\infty \frac{1}{k} \sin(\frac{x}{k})$ converges uniformly on [0,1] by the Weierstrass M-Test and then it uses it to prove something else. However, it does not show how to prove it and I get something very different.

This is what I have:

We have $f_n(x)=\frac{1}{k}\sin(\frac{x}{k})\le\frac{1}{k}=M_k$

But, $\sum\limits_{k=1}^\infty M_k=\sum\limits_{k=1}^\infty\frac{1}{k}$ diverges since it is a harmonic series.

I would really appreciate it if someone could help me prove how the series converges uniformly... Thanks!

  • $\begingroup$ On this interval, which doesn't include $\pi/2$, you can have much better approximations on the sine term. $\endgroup$
    – davidlowryduda
    Jun 29, 2014 at 14:55

2 Answers 2


Weierstrass M-test states: IF there is $M_k$ such that $|f_k(x)|\le\ M_k $ for all $k,x$ and $\sum\limits_{k=1}^\infty\ M_k$ converges, then $\sum\limits_{k=1}^\infty\ f_k(x)$ converges uniformly.

In particular, what you just have shown tells nothing about uniform convergence of the series.. since you just found 'a' sequence always bigger than f, whose series does not converge.

If you think about it, the term $\sin(\frac{x}{k})$ provides some additional speed of convergence to 0 when mutiplied to $\frac{1}{k}$ so overall the term converges to 0 faster than $\frac{1}{k}$ alone, and so the series does converge and does so uniformly

So obvious thing to try is to bound the term $\sin(\frac{x}{k})$ in some way (in your case you bounded it by 1).

Notice that on $[0,\infty)$ we have the following:

$g(x)=x-\sin(x) \implies g'(x)=1-\cos(x) \implies g'(x)\ge0$

and $g(0)=0$, hence $g(x)\ge0 \implies x\ge\sin(x)$

The above inequality can be obtained without using differentiation, if you'd like, using power series (which has infinite radius of convergence) for $\sin(x)$ as well.

Then for any $x\in[0,1]$

$$ \left|\frac{1}{k}\sin\left(\frac{x}{k}\right)\right|=\frac{1}{k}\sin\left(\frac{x}{k}\right)\le\frac{x}{k^2}\le\frac{1}{k^2}=M_k$$

and $\sum\limits_{k=1}^\infty M_k$ converges.


Use the well-known inequality (that is a direct consequence of the Mean Value Theorem): $$\forall x\in\mathbb{R},\ \lvert\sin x\rvert\leq\lvert x\rvert.$$ With your functions $f_n$ defined as: $$\forall n\in\mathbb{N}^*,\ \forall x\in[0,1],\ f_n(x)=\frac1n\sin\left(\frac xn\right),$$ we have: $$\lVert f_n\rVert_{\infty,[0,1]}\leq\frac1{n^2}$$ which is the general term of a convergent series (Riemann with $2>1$), hence the $M$-test applies, hence your series of functions converges uniformly on $[0,1]$.

More generally, with this method, you can prove that your series of functions converges uniformly on every bounded subset of $\mathbb{R}$.


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