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Suppose $\sum a_n $ is convergent and $a_n > 0$ for all $n$, does it follow that $\sum \left( \frac{ 1 + \sin (a_n) }{2} \right)^n $ is convergent??

yes

Since $\sum a_n$ is convergent, then $\sum \sin a_n$ is convergent so $\lim \sin a_n = 0 $ so using root test we know

$$ \sqrt[n]{ \left( \frac{ 1 + \sin (a_n) }{2} \right)^n } \to \frac{1 + 0}{2} = \frac{1}{2} < 1$$

so series converges by the root test. IS this correct?

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The argument is correct, although I would just say that when $a_n> 0$ and $\sum a_n$ is finite then $\lim a_n = 0$ whence $\lim \sin a_n=0$ and then use the root test.

This is probably not of your worry but in fact if there is $0<r<\pi/2$ so that $|a_n| < r$ for $n$ large enough the series converges already (for the same reason).

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Your route is correct, as a direct application of the Root Test.

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Yes that is a correct application of the Root Test. But you do not need the intermediate claim that the sum of the sines converges (which is true). Just put $a_n->0$ directly into the radical expression and you get the same result.

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