Let $f: (0,1) \rightarrow \mathbb{R}$ be increasing on $(0,1)$. If the improper integral (of second kind) $\int_0^1 f(t)dt$ converges ($f$ is unbounded at $0$ and at $1$), then show that the sequence $ (x_n)$ converges to $\int_0^1 f(t)dt$ where $x_n = \int_{\frac{1}{n}}^1 f(t)dt$.
My Approach: Fix $\epsilon > 0$, we need to find $N > 0$ such that:
$ n \geq N \Rightarrow \vert \int_{\frac{1}{n}}^1 f(t)dt - \int_0^1 f(t)dt \vert < \epsilon \\ \Leftrightarrow \vert \int_{0}^\frac{1}{n} f(t)dt \vert < \epsilon $
But since $f$ is unbounded as $x \rightarrow 0^+$, I don't know how to bound $\vert \int_{0}^\frac{1}{n} f(t)dt \vert $. How do I show this?
Use the definition of the improper integral. You do not need to treat the upper limit. It is $given$ that the improper integral converges; i.e., $x=\int^1_0f(t)dt$ is a number. Moreover, fixing $n$ for the moment, it is clear that $x_n=\int^1_{1/n}f(t)dt$ is well-defined because it is $\lim_{r\to 1} \int^r_{1/n}f(t)dt$ which exists by definition of the improper integral.
Now, you are asked to show that $\int_{1/n}^1 f(t)dt\to \int_0^1 f(t)dt$ as $n\to \infty.$ But again, this follows immediately from the definition of convergence of the improper integral.