[proofplan]
We establish integral comparison inequalities by bounding $f$ on each unit interval $[n-1, n]$ using monotonicity. Summing these bounds yields $\sum_{n=2}^N f(n) \le \int_1^N f(x)\,d\mathcal{L}^1(x) \le \sum_{n=1}^{N-1} f(n)$, from which both the convergence equivalence and the existence of the Euler-type limit follow. The limit exists because the partial-sum-minus-integral sequence is decreasing and bounded below.
[/proofplan]
[step:Establish the comparison inequalities from monotonicity]
Since $f$ is positive and decreasing, for each integer $n \ge 2$ and $x \in [n-1, n]$ we have $f(n) \le f(x) \le f(n-1)$. Integrating over $[n-1, n]$:
\begin{align*}
f(n) \le \int_{n-1}^{n} f(x)\,d\mathcal{L}^1(x) \le f(n-1).
\end{align*}
Summing from $n = 2$ to $N$:
\begin{align*}
\sum_{n=2}^{N} f(n) \le \int_1^N f(x)\,d\mathcal{L}^1(x) \le \sum_{n=1}^{N-1} f(n).
\end{align*}
Adding $f(1)$ to the left inequality and rearranging:
\begin{align*}
\sum_{n=1}^{N} f(n) - \int_1^N f(x)\,d\mathcal{L}^1(x) \le f(1).
\end{align*}
The right inequality gives $\int_1^N f(x)\,d\mathcal{L}^1(x) \le \sum_{n=1}^{N-1} f(n) \le \sum_{n=1}^{N} f(n)$, so
\begin{align*}
0 \le \sum_{n=1}^{N} f(n) - \int_1^N f(x)\,d\mathcal{L}^1(x) \le f(1).
\end{align*}
[guided]
Since $f$ is positive and decreasing, for each $x \in [n-1, n]$ we have $f(n) \le f(x) \le f(n-1)$. Integrating over $[n-1, n]$:
\begin{align*}
f(n) \le \int_{n-1}^{n} f(x)\,d\mathcal{L}^1(x) \le f(n-1).
\end{align*}
The left inequality $\sum_{n=2}^N f(n) \le \int_1^N f(x)\,d\mathcal{L}^1(x)$ is obtained by summing the left bound from $n = 2$ to $N$ and using additivity of the [integral](/page/Integral):
\begin{align*}
\sum_{n=2}^N f(n) \le \sum_{n=2}^N \int_{n-1}^n f(x)\,d\mathcal{L}^1(x) = \int_1^N f(x)\,d\mathcal{L}^1(x).
\end{align*}
The right inequality $\int_1^N f(x)\,d\mathcal{L}^1(x) \le \sum_{n=1}^{N-1} f(n)$ is obtained by summing the right bound from $n = 2$ to $N$:
\begin{align*}
\int_1^N f(x)\,d\mathcal{L}^1(x) = \sum_{n=2}^N \int_{n-1}^n f(x)\,d\mathcal{L}^1(x) \le \sum_{n=2}^N f(n-1) = \sum_{n=1}^{N-1} f(n).
\end{align*}
The non-negativity $\sum_{n=1}^N f(n) - \int_1^N f(x)\,d\mathcal{L}^1(x) \ge 0$ follows from the right inequality: $\int_1^N f(x)\,d\mathcal{L}^1(x) \le \sum_{n=1}^{N-1} f(n) \le \sum_{n=1}^N f(n)$ (since $f(N) > 0$). The upper bound $\sum_{n=1}^{N} f(n) - \int_1^N f(x)\,d\mathcal{L}^1(x) \le f(1)$ follows from the left inequality by adding $f(1)$ to both sides and rearranging.
[/guided]
[/step]
[step:Show the partial-sum-minus-integral sequence is decreasing and convergent]
Define $\phi: \mathbb{N} \to \mathbb{R}$ by $\phi(N) = \sum_{n=1}^{N} f(n) - \int_1^N f(x)\,d\mathcal{L}^1(x)$. Then
\begin{align*}
\phi(N) - \phi(N-1) = f(N) - \int_{N-1}^{N} f(x)\,d\mathcal{L}^1(x) \le f(N) - f(N) = 0,
\end{align*}
where the inequality uses $\int_{N-1}^N f(x)\,d\mathcal{L}^1(x) \ge f(N)$ (from the left bound in the previous step). Therefore $\phi$ is decreasing.
Since $\phi$ is decreasing and bounded below by $0$ (from the previous step), $\phi$ converges. The limit satisfies
\begin{align*}
0 \le \lim_{N \to \infty} \phi(N) \le \phi(1) = f(1).
\end{align*}
[/step]
[step:Deduce the equivalence of convergence]
From the comparison inequalities:
\begin{align*}
\sum_{n=1}^{N} f(n) \le f(1) + \int_1^N f(x)\,d\mathcal{L}^1(x), \qquad \int_1^N f(x)\,d\mathcal{L}^1(x) \le \sum_{n=1}^{N-1} f(n).
\end{align*}
If $\int_1^\infty f(x)\,d\mathcal{L}^1(x)$ converges, the [first inequality](/theorems/2897) shows $\sum_{n=1}^N f(n) \le f(1) + \int_1^\infty f(x)\,d\mathcal{L}^1(x)$ for all $N$. Since the partial sums are increasing (each $f(n) > 0$) and bounded above, the series $\sum f(n)$ converges.
Conversely, if $\sum f(n)$ converges, the [second inequality](/theorems/2899) shows $\int_1^N f(x)\,d\mathcal{L}^1(x) \le \sum_{n=1}^\infty f(n)$ for all $N$. Since $N \mapsto \int_1^N f(x)\,d\mathcal{L}^1(x)$ is increasing (because $f > 0$) and bounded above, $\int_1^\infty f(x)\,d\mathcal{L}^1(x)$ converges.
[/step]
