show that the series ∑∞n=1 sin n theta/n does not converge uniformly on the interval ]0,2π[.
"\\sum_{n=1}^{\\infty} \\frac{sin n\\theta}{n}\n, \\theta\\in]0,2\\pi[.\\\\\nThen,\\\\\nf_n=\\frac{sin n\\theta}{n}\n\\to0as n\\to \\infty\\\\\nf'_n=cosn\\theta\\\\\nf'_n(\\pi)=(-1)^n\\text{ does not converge as } n\\to \\infty\n\\\\\n\\text{Thus, the given series is not uniformly covergent.}\\\\"
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