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I am trying to show that the series $$\sum _{n=1}^{\infty} \sin n$$ diverges but has bounded partial sums.

Plugging in some terms we see that,

$$\sum _{n=1}^{\infty} \sin n = \sin1 +\sin2 +\sin3+...+\sin n$$

My idea is to try and use $$e^{i\theta} = \cos\theta+i\sin\theta$$ $$e^{in}=\cos n+\sin n$$ $$\sin n = Im(e^{in})$$

But how can I use this to show that a finite geometric series won't converge to anything, therefore diverge, but is bounded?

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    I strongly disagree with such widespread terminology: how can a sequence be *bounded* and *divergent* at the same time? In Italy (and perhaps in France, too), *divergent* is not the opposite of *convergent*, it means that $|a_n|\to +\infty$. It makes much more sense. – 2017-02-02
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    @JackD'Aurizio do you have a possible answer using my logic? – 2017-02-02
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    @fr14: you question has already been asked on MSE many times. The sequence cannot be convergent because $\sin(n)$ does not converge to $0$ as $n\to+\infty$, and is bounded since $\sin(n)$ is the imaginary part of $e^{in}$, so your partial sums are bounded like the imaginary part of bounded "geometric sums". – 2017-02-02
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    okay thanks, I will try and understand this answer – 2017-02-02

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$\sin n$ doesn't tend to $0$ while $n\to\infty$, therefore the sum of the series doesn't converge. To show the first, just note that there exists a constant $\varepsilon>0$ such that one of $|\sin n|, |\sin(n+1)|$ is greater than $\varepsilon$.

But $$\sum_{n=0}^Ne^{in}=\frac{e^{i(N+1)}-1}{e^i-1}$$ The denominator is constant in $N$, and the numerator is bounded by 2, so this series is bounded. Then its imaginary part is also bounded by the same constant.

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    To show the first, just note that there exists a constant... from this sentence on does it show that sin(n) is bounded? . – 2017-02-02
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    $sin(n)$ is bounded by 1, but this seems to be not connected with the problem. I meant, when $sin(n)$ is close to zero, then $sin(n+1)$ is close to $\pm sin(1)$ and so is not close to zero. – 2017-02-02