How do we discuss the convergence of the following series
i)$\sum _{n=1}^\infty (\sin n)^n $
ii)$\sum _{n=1}^\infty (\cos n)^n $
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2They both diverge, as the terms don't go to zero. – davidlowryduda Apr 12 '14 at 07:58
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Can you make a multiple of $\pi $ by running through the natural numbers (mod $2\pi$)? – Ellya Apr 12 '14 at 08:04
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(Integer multiple) – Ellya Apr 12 '14 at 08:29
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2@mixedmath: I'm sure you're right, but it's not obvious! – TonyK Apr 12 '14 at 09:08
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HINT : Show that there exists a sequence of integers $(p_n)$ and a sequence of integers $(q_n)$ such that $|\pi-\frac{p_n}{q_n}|<\frac{1}{q_n^2}$ for ii)
And, for i) Show that there exists a sequence of integers $(p_n)$ and a sequence of integers $(q_n)$ chosen so that all the terms of ${q_n}$ are odd, such that $|\frac{\pi}{2}-\frac{p_n}{q_n}|<\frac{1}{q_n^2}$
Then, $$\vert \cos(p_n)\vert=\vert \cos(\pi q_n-p_n)\vert>\cos(\frac{1}{q_n})=1-2\sin^2(\frac{1}{2q_n})>1-\frac{1}{2q_n^2}$$ and, $$\vert \cos(p_n)\vert^{p_n}>\bigr(1-\frac{1}{2q_n^2}\bigl)^{p_n}> 1-\frac{p_n}{2q_n^2} $$
Therefore, the sequence $\{(cos(p_n))^{p_n}\}$and $\{ \cos(n)^n\}$ doesn't converge to $0$. It follows that ii) diverge.
For i) It's the same sketch so I leave it to you.
