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Let $\{a_n \}$ be a decreasing sequence of positive real numbers. Prove that if the series

$$\sum_{n=0}^{\infty} a_n \sin (nx)$$

converges uniformly on $\mathbb R$ then $na_n \rightarrow 0$.

I find difficulty to prove this. Please help me.

Thank you in advance.

Arnab Chattopadhyay.
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    How does the convergence of $\sum_{n=0}^{\infty} a_n \sin (nx)$ imply convergence of $\sum_{n=0}^{\infty} a_n$? – Arnab Chattopadhyay. Dec 02 '17 at 18:06
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    Use the uniform convergence of the series. Since the series is uniformly convergent so for a given $\epsilon>0$ $\exists$ $N \in \mathbb N$ such that $\forall n \ge N$ we have $|\sum_{k=n}^{2n-1} a_k \sin (kx)| < \epsilon$. Now take $x=\frac {1} {2n}$ and observe that $\sin (\frac {1} {2}) \le \sin (kx) \le \sin 1$. Then you will have $\epsilon > |\sum_{k=n}^{2n-1} a_k \sin (kx)| = \sum_{k=n}^{2n-1} a_k \sin (kx) \ge \sum_{k=n}^{2n-1} a_{2n} \sin (\frac {1} {2}) = 2na_{2n} ({\frac {1} {2}} \sin(\frac {1} {2}))$ $\implies$ $2na_{2n} \rightarrow 0$. Similarly you can prove for the odd terms.ok! –  Dec 03 '17 at 08:02
  • very nice answer! – Arnab Chattopadhyay. Dec 03 '17 at 08:16

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