# Show that $f(x)=\sum_{k=1}^\infty \frac{1}{k}\sin(\frac{x}{k+1})$ converges.

Exercise: Show that $$f(x)=\sum_{k=1}^\infty \frac{1}{k}\sin\left(\frac{x}{k+1}\right)$$ converges, pointwise on $\mathbb{R}$ and uniformly on each bounded interval in $\mathbb{R}$, to a differentiable function $f$ which satisfies $$|f(x)|\leq |x| \text{ and } |f'(x)|\leq 1$$ for all $x\in \mathbb{R}$.

Hint: Dominate, then telescope.

I am unsure how to start this proof. By definition, I know that I need to show that the sequence with the terms $$s_n(x)=\sum_{k=1}^n \frac{1}{k}\sin\left(\frac{x}{k+1}\right)$$ converges pointwise on $\mathbb{R}$ and uniformly on a bounded interval in $\mathbb{R}$, but I am unsure how to show these facts.

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Let's denote by $$f_n(x)=\frac{1}{n}\sin(\frac{x}{n+1}),$$ so we have $$f(x)=\sum_{n=1}^\infty f_n(x).$$

It's clear that $f$ is defined at $0$ and $$f_n(x)\sim\frac{x}{n^2},\forall x\neq0$$ then we have pointwise convergence by comparaison with the Riemann series.

Now, let $[a,b]$ a bounded interval in $\mathbb{R}$. We have $$|f_n(x)|\leq|\frac{x}{n^2}|\leq \frac{\max(|a|,|b|)}{n^2},$$ so we have normal convergence which implies the uniform convergence of the series on $[a,b]$.

Moreover, from $$|f'_n(x)|=\left|\frac{\cos(\frac{x}{n+1})}{n(n+1)}\right|\leq\frac{1}{n^2},$$ hence, we find the uniform convergence of the series $\sum_nf'_n(x)$ on $\mathbb{R}$ which prove that $f$ is differentiable and $$f'(x)=\sum_{n=1}^\infty f'_n(x).$$ Finally, we have these two inequality $$|f(x)|\leq\sum_{n=1}^\infty \frac{1}{n}|\sin\frac{x}{n+1}|\leq|x|\sum_{n=1}^\infty(\frac{1}{n}-\frac{1}{n+1})=|x|,$$

and $$|f'(x)|\leq\sum_{n=1}^\infty|f'_n(x)|\leq\sum_{n=1}^\infty(\frac{1}{n}-\frac{1}{n+1})=1.$$

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Compare with $$\sum_{k=1}^\infty \dfrac{x}{k(k+1)}.$$
Since $|\sin(\theta)|\leq |\theta|$ for all $\theta\in \mathbb{R}$, $\left|\frac{1}{k}\sin\left(\frac{x}{k(k+1)}\right)\right|\leq \left|\frac{x}{k(k+1)}\right|$. Then we show that the series $\sum_{k=1}^\infty \frac{x}{k(k+1)}$ converges and so the original series converges uniformly by the Weierstrass M-Test? –  kaiserphellos Mar 7 '13 at 10:33