Jtdylktuy

Theorem: Assume there is an $Ninmathbb{N}$ so that $f:[N,infty)to mathbb{R}$ is non-negative, continuous and decreasing. Define $a_n=f(n)$ for $nin mathbb{N}$ with $ngeq N$. 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
$$
a_{n+1}=f(n+1)leq int_{n}^{n+1}f(x),mathrm{d}xleq f(n)=a_n
$$
for all $ngeq N$. Defining $b_n=int_{n}^{n+1}f(x),mathrm{d}x$ for $ngeq N$, 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
$$
int_{N}^{infty}f(x),dx:=lim_{Atoinfty}int_{N}^{A}f(x),dx
$$
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 $Mgeq N$. 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 $Mgeq N$. I am wondering, if this implies $int_{N}^{A+1}f(x),mathrm{d}x<K$ for all real numbers $Ageq N$ ...

(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
$$
int_N^{lceil A+1rceil} f(x), dx = int_N^{A+1} f(x), dx + int_{A+1}^{lceil A+1rceil} f(x), dx geq K + int_{A+1}^{lceil A+1rceil} f(x), dx geq K.
$$
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 $xgeq M$ that
$$
L - int_N^{x+1} f(x),dx leq L - int_N^{M+1} f(x),dx < epsilon.
$$