Theorem: Assume there is an $N\in\mathbb{N}$ so that $f:[N,\infty)\to \mathbb{R}$ is non-negative, continuous and decreasing. Define $a_n=f(n)$ for $n\in \mathbb{N}$ with $n\geq N$. Then, $\sum_{n=N}^{\infty}a_n$ converges if and only if $\int_{N}^{\infty}f(x)\,dx$ converges.
Proof: Since $f$ is decreasing, we find that $$ a_{n+1}=f(n+1)\leq \int_{n}^{n+1}f(x)\,\mathrm{d}x\leq f(n)=a_n $$ for all $n\geq N$. Defining $b_n=\int_{n}^{n+1}f(x)\,\mathrm{d}x$ for $n\geq N$, it follows from the Comparison Test that $\sum_{n=N}^{\infty}a_n$ converges if and only if $\sum_{n=N}^{\infty}b_n$ converges.
Question: Is $$ \sum_{n=N}^{\infty}b_n \textrm{ converges}\implies \int_{N}^{\infty}f(x)\,dx \textrm{ converges}? $$
I am asking this, because of the definition $$ \int_{N}^{\infty}f(x)\,dx:=\lim_{A\to\infty}\int_{N}^{A}f(x)\,dx $$ where $A$ is a real number.
Edit: Found something in LINK, on the last page, in particular Lemma 3. Note that $$ \sum_{n=N}^{M}b_n=\int_{N}^{M+1}f(x)\,\mathrm{d}x $$ for all integers $M\geq N$. So, if $\sum_{n=N}^{\infty}b_n$ converges, the partial sums of it is bounded above by some positive number $K$. Therefore $\int_{N}^{M+1}f(x)\,\mathrm{d}x<K$ for all $M\geq N$. I am wondering, if this implies $\int_{N}^{A+1}f(x)\,\mathrm{d}x<K$ for all real numbers $A\geq N$ ...
(1) For your second question in the edit: suppose there is a real number $A \geq N$ such that $$ \int_N^{A+1} f(x)\, dx \geq K. $$ Since $f$ is non-negative, the integral of $f$ from $A+1$ to $\lceil A+1 \rceil$ is non-negative as well. Therefore $$ \int_N^{\lceil A+1\rceil} f(x)\, dx = \int_N^{A+1} f(x)\, dx + \int_{A+1}^{\lceil A+1\rceil} f(x)\, dx \geq K + \int_{A+1}^{\lceil A+1\rceil} f(x)\, dx \geq K. $$ But since $\lceil A+1 \rceil$ is a natural number, we have a contradiction.
(2) Now let's prove your original question. Assume the series converges to $L$. The argument in paragraph (1) shows that $L- \int_N^{x+1}f(x)\,dx \geq 0$ for every real $x$. Take $\epsilon > 0$ arbitrary. Since the series converges, there is a natural number $M \geq N$ such that $$ L - \int_N^{M+1} f(x)\,dx = L- \sum_{n=N}^M b_n < \epsilon. $$ Note that the function $\int_N^{x+1} f(x)\,dx$ is increasing since $f$ is non-negative. Therefore we have for every $x\geq M$ that $$ L - \int_N^{x+1} f(x)\,dx \leq L - \int_N^{M+1} f(x)\,dx < \epsilon. $$