Proving Integral Test?

Question

Assume that $f(x) \geq 0$ and that $f$ decreases monotonically on $[1, \infty]$.
Prove $\int_{1}^{\infty} f(x)dx$ converges iff $\sum_{n=1}^{\infty} f(n)$ converges.

My proof:

If $f$ is non-negative and monotonically decreasing, it is integrable on any finite interval. Consider a partition of $[1,t]$ consisting of the $t$ points $1,2,...,t$. Since the upper Riemann sum for the partition is $\sum_{n=1}^{t-1} f(n)$ and the lower Riemann sum is $\sum_{n=2}^{t} f(n)$, it clearly follows that $$\sum_{n=2}^{t} f(n) \leq \int_{1}^{t} f(x)dx \leq \sum_{n=1}^{t-1} f(n).$$ Hence we have that $$\sum_{n=1}^{\infty} f(n) -f(1) \leq \lim_{t \rightarrow \infty} \int_{1}^{t} f(x)dx \leq \sum_{n=1}^{\infty} f(n).$$ It follows that the seqeuence of partial sums of the series $\sum_{n=1}^{\infty} f(n)$ is bounded if and only if $\lim_{t \rightarrow \infty} \int_{1}^{t} f(x)dx$ is bounded, hence the limits either both exist or both are infinite. Thus the result follows.

Something feels off... would the above proof be sufficiently rigorous?

Answer 1

Apart from one missing detail - the integrability of $f$ - this looks fine to me. You can deduce integrability fairly easily, and may already have established that result. As pointed out in the comments, you could expand on the "converge or diverge together" comment as well.

Answer 2

First review what a Riemann sum is and what the definition of an improper integral is. Your sums are not Riemann sums. Then, choose an arbitrary partition $P$ of your interval $[0,t]$, find a relationship between the upper sums $U(P)$, the lower sums $L(P)$ and $f$ and apply the squeeze theorem or use an $\epsilon - \delta $ argument by estimating the difference $U(P)-L(P)$. Finally, take the limit as $t\rightarrow \infty $.