Le be $f:[1,\infty[ \to [0,\infty[$ and $f$ is monotonously decreasing. Further, $f$ is Riemann integrable on each of the domains $[1,\beta]$ where $\beta \in ]1,\infty[$.
Prove the following equivalence:
$\int^{\infty}_1 f(x) dx$ is convergent (the improper integral exists) $\Leftrightarrow$ the series $\sum^{\infty}_{k=1}f(k)$ is convergent.
My approach:
Firstly, I assume that $\sum\limits_{k=1}^{\infty} f(k)$ is convergent.
So we know from lecture that for an arbitrary $\epsilon>0$ there exists a $n_0\in\mathbb{N}$ such that for all $n>m>n_0$, where $m,n\in \mathbb{N}$, we get: $\sum\limits_{k=m}^n f(k)<\epsilon$.
Let be $\beta_1,\beta_2 \in \mathbb{R}$ arbitrary chosen with $\beta_2>\beta_1>n_0+1$. We define a partition of $[\beta_1,\beta_2]$ as follows: $P':=\{\beta_1,\lfloor\beta_1\rfloor+1, \lfloor\beta_1\rfloor+2, ..., \lfloor\beta_2\rfloor,\beta_2\}$. So if $(\beta_2-\beta_1)< 1$ then $P'$ simply becomces $P'=\{\beta_1, \beta_2\}$. As $f$ is Riemann integrable on $[\beta_1,\beta_2]$ we can write:
$\int^{\beta_2}_{\beta_1}f(x)dx=\inf\{U(f,P)|$ P is a partition of $[\beta_1,\beta_2]\}$. (Where $U$ is an upper Darboux sum)
As $P'$ has a lower refinement than the $\inf$ and $f$ is monotonously decreasing we can use the following upper boundary:
$...\leq U(f,P')= f(\beta_1)(\lfloor\beta_1\rfloor+1 -\beta_1)+f(\lfloor\beta_1\rfloor+1)+f(\lfloor\beta_1\rfloor+2)+...+f(\lfloor\beta_2\rfloor)(\beta_2 -\lfloor\beta_2\rfloor)\leq \sum\limits_{k=\lfloor\beta_1\rfloor}^{\lfloor\beta_2\rfloor}f(k)<\epsilon$.
The last two inequalities hold because $f$ is monotonously decreasing and $\lfloor\beta_2\rfloor> \lfloor\beta_1\rfloor>n_0$ and $ \lfloor\beta_2\rfloor, \lfloor\beta_1\rfloor\in\mathbb{N}$. So we have found a $\beta_0$, namely $\beta_0:=n_0+1$, such that for an arbitrary $\epsilon>0$ and two arbitrary $\beta_1, \beta_2$ with $\beta_2> \beta_1>\beta_0$ we have $\int^{\beta_2}_{\beta_1}f(x)dx<\epsilon$. This proves the convergence of $\int^{\infty}_1 f(x) dx$ (in our lecture we call this the Cauchy criterion for improper integrals).
Now I assume that $\int^{\infty}_1 f(x) dx$ is convergent.
So we know that if we take an arbitrary $\epsilon >0$ there exists a $\beta_0$ such that for all $\beta_2>\beta_1>\beta_0$ we have:
$\int^{\beta_2}_{\beta_1} f(x) dx<\epsilon$. We can simply choose $\beta_2,\beta_1 \in \mathbb{N}$. We define a partition $P':=\{\beta_1, \beta_1+1, \beta_1+2, ..., \beta_2\}$ for the interval $[\beta_1,\beta_2]$. As $\int^{\beta_2}_{\beta_1} f(x) dx$ is Riemann integrable we get:
$\epsilon>\int^{\beta_2}_{\beta_1} f(x) dx= sup\{L(f,P)|$ P is a partition of $[\beta_1,\beta_2]\}$ (where $L$ is a lower Darboux sum) $...\geq L(f,P')= f(\beta_1)+f(\beta_1+1)+f(\beta_1+2)+...+f(\beta_2)= \sum\limits_{k=\beta_1}^{\beta_2}f(k)$. By means of the Cauchy criterion (in the context of sequences and series) the series $\sum\limits_{k=1}^{\infty} f(k)$ is convergent.
Is this approach correct? Any feedback is appreciated :)