Krdytkyu

Consider the points $p_n=(cos n,sin n),$ $ n=1,2,dots $ on the unit circle. The $p_n$ march around the circle infinitely many times in steps of of arc length $1.$ Let $A$ denote the arc on the unit circle between the points $(1/sqrt 2,1/sqrt 2)$ and $(-1/sqrt 2,1/sqrt 2).$ The length of this arc is $pi/2>1.$ It is impossible for our sequence $p_n$ to jump over $A,$ as the $p_n$ are marching around in steps of arc length $1.$ Thus $p_nin A$ for infinitely many $n.$ For each such $n,$ we have $sin n ge 1/sqrt 2.$ It follows that $(e^nsin n)/n ge e^n/(sqrt 2cdot n)$ for infinitely many $n.$ Thus the terms of our series do not approach $0,$ which shows the series diverges.

If $|sin(n),|lesinleft(frac12right)$, then $|cos(n),|gecosleft(frac12right)$ and
$$
begin{align}
|sin(n+1),|
&=|cos(n)sin(1)+sin(n)cos(1),|\
&gecosleft(tfrac12right)sin(1)-sinleft(tfrac12right)cos(1)\
&=sinleft(tfrac12right)
end{align}
$$
This means that either $|sin(n),|gesinleft(frac12right)$ or $|sin(n+1),|gesinleft(frac12right)$.

Since $e^ngt1+n$, we know that $frac{e^n}ngt1$.

Therefore, either $left|frac{e^nsin(n)}nright|gesinleft(frac12right)$ or $left|frac{e^{n+1}sin(n+1)}{n+1}right|gesinleft(frac12right)$. That is, the terms do not tend to $0$.

Take the root test. We have $limsup_{ntoinfty}(|e^nsin(n)|)^{1/n} = e> 1$ so it diverges.

$$ lim_{n rightarrow infty} frac{e^nsin n}{n} neq 0 $$
So the series diverges using the nth term test for divergence.

It suffices to show that $sin(n)$ fails to approach $0$, which means that your series fails to converge by the $n$th term test (that is, the sequence of summands fails to approach $0$).

Some proofs of that $sin(n)$ fails to converge to $0$ are given here.

I would suspect, however, that if you are learning about testing for convergence/divergence of series for the first time, you would not be expected to prove such a result. If it comes up, I would state without proof that $sin(n)$ does not have a limit as $nto infty$, just as the function $f(x) = sin(x)$ fails to have a limit as $x to infty$.

The series does not converge because the general term does not tend to zero. To prove this, it is enough to show that $sin n$ does not converge. Here is an elementary way to do it: suppose that $sin nto xin mathbb R$.

Then, of course,

$sin 2nto x$ and $sin (2(n+1))to x.$

And so

$sin nto xRightarrow cos 2nto 1-2x^2Rightarrow cos (2(n+1))to 1-2x^2$

But now,

$sin 2=sin(2(n + 1) - 2n)=sin(2(n + 1))cdotcos 2n-sin 2ncdot cos (2(n+1))to 0 $, which is absurd.

Assuming $frac{e^n sin n}{n}to 0$ as $nto +infty$ we have $left|sin(n)right|leq frac{n}{2e^n}leqsinfrac{1}{2^n}$ for any $n$ sufficiently large, hence $left|cos nright|geq cosfrac{1}{2^n}$. On the other hand such assumptions lead to
$$ left|sin(n+1)right|geq left|left|sin(n)right|cos(1)-left|cos(n)right|sin(1)right|geq frac{4}{5} $$
for any $n$ sufficiently large, contradiction.