Does there exist a constant $M$ st $S(n)=\sum_{k=0}^n\left(e^{\sin k}-I_0(1)\right)\le M$? Basically the title. Is the sequence
$$a_n=\sum_{k=0}^n\left(e^{\sin k}-\frac1{2\pi}\int_{-\pi}^{\pi}e^{\sin x}dx\right)$$
bounded? Intuitively, it sort of makes sense because the integral is sort of the average value of $e^{\sin x}$, so I wouldn't expect it to blow up to infinity.
I'm not too knowledgeable about analysis, so correct me if I'm using wrong terminology.
Is $S(n)=\theta(1)$?
 A: Indeed $S(n)$ is bounded. Write
$$
\exp(\sin x)
= \exp\left(\frac{e^{i\theta}-e^{-i\theta}}{2i}\right)
= \sum_{a, b \geq 0} \frac{i^{b-a}}{a!b!2^{a+b}} \, e^{i(a-b)\theta}.
$$
Then
$$
I_0(1)
= \frac{1}{2\pi}\int_{-\pi}^{\pi} e^{\sin x} \, \mathrm{d}x
= \sum_{a, b \geq 0} \frac{1}{a!b!2^{a+b}} \mathbf{1}_{\{a=b\}},
$$
and so,
$$
\left| S(n) \right|
\leq \sum_{\substack{a, b \geq 0 \\ a \neq b}} \frac{1}{a!b!2^{a+b}} \left| \sum_{k=1} ^{n} e^{i(a-b)k} \right|
= \sum_{\substack{a, b \geq 0 \\ a \neq b}} \frac{1}{a!b!2^{a+b}} \frac{1}{\left|\sin\left(\frac{a-b}{2}\right)\right|}
$$
Now by using the fact that $\frac{1}{2\pi}$ has finite irrationality measure, there exist $c > 0$ and $\alpha > 0$ such that
$$ \left| \frac{q}{2\pi} - p \right| \geq c q^{-\alpha} $$
for any $p \in \mathbb{Z}$ and $q \in \mathbb{Z}_{>0}$. So, if we choose $p$ so that $\left| \frac{q}{2\pi} - p \right| \leq \frac{1}{2}$, then using the inequality $\left| \sin x \right| \geq \frac{2}{\pi} \left| x \right| $ which is valid for $-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$ and the above inequality,
$$ \left| \sin \left(\frac{q}{2}\right) \right|
= \left| \sin \left(\frac{q}{2} - \pi p\right) \right|
\geq \frac{2}{\pi}\left| \frac{q}{2} - \pi p \right|
\geq 2c q^{-\alpha}. $$
Therefore it follows that
$$ \left| S(n) \right| \leq \sum_{\substack{a, b \geq 0 \\ a \neq b}} \frac{1}{a!b!2^{a+b}} \frac{\left| a-b \right|^\alpha}{2c}, $$
and this upper converges. $\square$

The figure below demonstrates first $10^4$ values of $S(n)$, which also suggests that $S(n)$ should be bounded:

