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.

We can write the $Si$ function as 
$$\begin{align}
Si(n)&=\int_0^n \frac{\sin x}{x}dx\\
&=\int_0^1 \frac{\sin x}{x}dx+\int_1^n \frac{\sin x}{x}dx
\end{align}$$
The first integral on the right-hand side is obviously convergent.  Let's examine the second one more carefully.
Integrating by parts reveals
$$\begin{align}
\lim_{n\to \infty} \int_1^n \frac{\sin x}{x}dx&=\cos(1)+\lim_{n\to \infty}\left(-\int_1^n \frac{\cos x}{x^2}dx\right)
\end{align}$$
Since the integral on the right-hand side satisfies the inequality
$$\begin{align}
\left|-\int_1^n \frac{\cos x}{x^2}dx\right|&\le \int_1^n \left|\frac{\cos x}{x^2}\right|dx\\\\
&\le \int_1^n \frac{1}{x^2}dx\\\\
&=1-\frac1{n}
\end{align}$$
it is absolutely convergent.  Thus, the original integral, $Si(n)$ converges (not absolutely), as was to be shown!
