Proving convergence of an improper integral If $f$ is a non negative-valued continuous function with domain $[1,\infty)$ and $\lim\limits_{n \to \infty} \int_{1}^n f(x)\,dx$ exists, then prove the improper integral $\int_{1}^\infty f(x)\,dx$ converges. 

I know that for the first part to have a limit that exists, it must mean that it itself is convergent since it is monotone and bounded by the limit. Not sure where to go from there.
Let $\lim_{n\to \infty} \int_1^n f = L.$ For $x &gt; 0,$ we have by the nonnegativity of $f$ that

$$\tag 1 \int_1^{F(x)} f \le \int_1^{x} f \le \int_1^{F(x) + 1} f.$$

Here $F$ denotes the floor function. As $x\to \infty,$ both $F(x), F(x) + 1\to \infty$ through integer values. Hence both the left and right side of $(1)$ tend to $L$ as $x \to \infty.$ By the squeeze theorem, the middle term must also $\to L.$ That is the very definition of $\int_1^\infty f = L.$
Note that $\lim\limits_{n \to \infty} \int_{1}^n f(x)\,dx$ exists, hence, let $N=\lim\limits_{n \to \infty} \int_{1}^n f(x)\,dx$, for some real $N$. Also, since $f$ is continuous on the domain $[1,\infty)$, 

$$\int_{1}^\infty f(x)\,dx=\lim\limits_{n \to \infty} \int_{1}^n f(x)\,dx=N$$

In this case, $\int_{1}^\infty f(x)\,dx$ converges to $N$. 



Note The following definitions of improper integrals of Type 1:

$(a)$ If $\int_{a}^n f(x)\,dx$ exists for every number $n\ge$$a$, then $\int_{a}^\infty f(x)\,dx=\lim\limits_{n \to \infty} \int_{a}^n f(x)\,dx$ provided that this limit exists (as a finite number).

$(b)$ If $\int_{n}^b f(x)\,dx$ exists for every number $n\le$$b$, then $\int_{-\infty}^b f(x)\,dx=\lim\limits_{n \to -\infty} \int_{n}^b f(x)\,dx$ provided that this limit exists (as a finite number).

The improper integrals $\int_{a}^\infty f(x)\,dx$ and $\int_{-\infty}^b f(x)\,dx$ are called convergent if the corresponding limits exist and divergent if the limits do not exist.