Integral test

Step 1 (bound each term by a strip of the integral). Since $f(x)$ is decreasing, for every integer $n\ge1$ and $x\in[n,n+1]$ we have $f(n+1)\le f(x)\le f(n)$. Integrating over that unit interval gives $f(n+1)\le\int_n^{n+1}f(x)\,dx\le f(n)$.

Step 2 (sum the bounds). Summing the right-hand inequality for $n=1,\dots,N$ gives $\int_1^{N+1}f(x)\,dx\le\sum_{n=1}^{N}a_n=S_N$; summing the left-hand inequality for $n=1,\dots,N-1$ gives $S_N-a_1\le\int_1^{N}f(x)\,dx$. So the partial sum $S_N$ is sandwiched between two integrals over intervals that both grow like $[1,N]$.

Step 3 (convergence transfers one way). If $\int_1^{\infty} f(x)\,dx$ converges, then $\int_1^{N}f(x)\,dx$ is bounded above as $N\to\infty$, so by Step 2 the increasing sequence $S_N$ is bounded above, and a bounded increasing sequence of real numbers always converges (a form of the monotone convergence property of $\mathbb{R}$); hence $\sum_{n=1}^{\infty} a_n$ converges.

Step 4 (divergence transfers the other way). Conversely, if $\int_1^{\infty} f(x)\,dx$ diverges (grows without bound), then $\int_1^{N+1}f(x)\,dx\to\infty$ as $N\to\infty$, and by the first inequality in Step 2, $S_N$ is bounded below by a quantity tending to infinity, so $S_N$ also diverges to infinity. Combining Steps 3 and 4 gives the full 'if and only if'.