# How can I show this series converges uniformly?

Suppose $f(x) =\sum_{k=1}^\infty \dfrac{1}{k}\sin\left( \dfrac{x}{k+1}\right)$. Show that $f(x)$ converges uniformly on any closed, bounded interval $[a,b]$.

I used the Weierstrass M-Test, but I am not sure if the series I chose to compare it to is accurate. Let $f_k(x) = \dfrac{1}{k}\sin\left( \dfrac{x}{k+1}\right)$. Since $\sin\left( \dfrac{x}{k+1}\right)\leq\left( \dfrac{x}{k+1}\right)$, we can say that $f_k(x) \leq \dfrac{x}{k(k+1)}$ for all $x$. Now, we need to show that $\sum_{k=1}^\infty \dfrac{x}{k^2+k}$ converges. $$\sum_{k=1}^\infty \dfrac{x}{k^2+k} = x \sum_{k=1}^\infty \dfrac{1}{k^2+k} \leq x\sum_{k=1}^\infty \dfrac{1}{k^2}$$ Since $\sum_{k=1}^\infty \dfrac{1}{k^2}$ converges, our original series converges. Thus, we can apply the Weierstrass M-test with $M_k = \dfrac{x}{k^2+k}$ and we have that $f(x)$ converges uniformly.

Any suggestions/improvements would be greatly appreciated. Thanks!

• You almost got it right, but your $M_k$ are not allowed to depend on $x$. They have to bound $f(x)$ for all $x \in [a,b]$. So instead take $M_k = \frac{ \max \{ |a|,|b| \}}{k^2+k}$ Mar 27, 2015 at 2:37

Use the mean value theorem to show that $|\sin t| \le |t|$ for all $t\in \Bbb R$. Using this, note that on $[a,b]$,
$$\left|\frac{1}{k}\sin\frac{x}{k+1}\right| \le \frac{|x|}{k(k+1)} \le \frac{\max\{|a|, |b|\}}{k(k+1)}.$$
Since $\sum_{k = 1}^\infty \max\{|a|,|b|\}/[k(k+1)]$ converges, the Weierstrass $M$-test gives uniform convergence of your series in $[a,b]$.