Proof $\text{Si}(n) $ is convergent I am trying to prove that the sequence formed by the Si function, $\text{Si}(n) = \int_0^n \frac{\sin(u)}{u} \mathrm{d}u$, is convergent as $n\rightarrow \infty$. The only twist is the lower bound of the integral is 1 instead of zero. At first I looked at this as a sort of alternating series, and the sequence formed was that of the partial sums. I attempted to change the upper bound of the integral so that I was looking at integer multiples of $\pi$. I was then going to bound each "bump" with a rectangle with base pi, and height equal to the function value at the middle each interval. Shortly in I realized that the "bumps" were not symmetric about their centers, so I did not have a way to calculate the height of the function at these midpoints. I cannot think of another way to attack this problem. Any help appreciated.

Hint: Prove that
$$
\int_0^{2n\pi}\frac{\sin(x)}{x}\,\mathrm{d}x
$$
converges using the estimates
$$
\int_0^{2\pi}\left|\,\frac{\sin(x)}{x}\,\right|\,\mathrm{d}x\le2\pi
$$
and, for $n\ge2$,
$$
\begin{align}
\left|\,\int_{2(n-1)\pi}^{2n\pi}\frac{\sin(x)}{x}\,\mathrm{d}x\,\right|
&=\left|\,\int_{2(n-1)\pi}^{2n\pi}\left(\frac{\sin(x)}{x}-\frac{\sin(x)}{2n\pi}\right)\,\mathrm{d}x\,\right|\\
&\le\int_{2(n-1)\pi}^{2n\pi}\left|\,\frac{\sin(x)}{x}-\frac{\sin(x)}{2n\pi}\,\right|\,\mathrm{d}x\\
&\le\int_{2(n-1)\pi}^{2n\pi}\left|\,\frac1{2n(n-1)\pi}\,\right|\,\mathrm{d}x\\
&=\frac1{n-1}-\frac1n
\end{align}
$$
Then prove that
$$
\int_{2n\pi}^{2n\pi+s}\left|\,\frac{\sin(x)}{x}\,\right|\,\mathrm{d}x\le\frac1n
$$
for $0\le s\le2\pi$
