Hello everyone! While doing some problems I came upon this one. What is disconcerting is that I have found (in general) that the more useful a theorem is the more difficult it is to prove. So I would appreciate if someone would tell me if this looks correct?

Terminology:

$\displaystyle P_n$ a partition of $\displaystyle [a,b]$ containing $\displaystyle n$ points with $\displaystyle x_1=a$.

$\displaystyle \wp$ is the set of all partitions of $\displaystyle [a,b]$

$\displaystyle \Delta \alpha_i=\alpha(x_{i+1})-\alpha(x_{i-1})~~\alpha\in\uparrow$

$\displaystyle M_i=\sup_{x_{i}\leqslant x\leqslant x_{i+1}}f(x)$

$\displaystyle m_i=\inf_{x_{i}\leqslant x\leqslant x_{i+1}}f(x)$

Taking $\displaystyle \alpha(x)=x$

$\displaystyle U\left(P_n,f\right)=\sum_{i=1}^{n-1}M_i\cdot\Delta x_i$

$\displaystyle L\left(P_n,f\right)=\sum_{i=1}^{n-1}m_i\cdot\Delta x_i$

Now suppose that $\displaystyle f\in\mathcal{R}$ ($\displaystyle f$ is Riemann integrable) on $\displaystyle [a,b]$ then $\displaystyle \int_a^b f ~dx=\inf_{P_n\in\wp}U\left(P_n,f\right)=\sup_{P_n\ in\wp}L\left(P_n,f\right)$

Note: by how we defined $\displaystyle \int_a^b f~dx$ it follows that for all partitions $\displaystyle L\left(P_n,f\right)\leqslant \int_a^b f~dx\leqslant U\left(P_n,f\right)$

Ok now onto the question

Question: Suppose that $\displaystyle f(x)$ is a positive, monotonically decreasing function, prove that $\displaystyle \sum_{x=1}^{\infty}f(x)\text{ converges}\Longleftrightarrow\int_1^{\infty}f~dx\t ext{ converges}$

Answer: Part one $\displaystyle \sum_{x=1}^{\infty}f(x)\text{ converges}\implies \int_1^{\infty} f~dx\text{ converges}$. Consider the interval $\displaystyle [1,b]$ with $\displaystyle b\in\left\{2,3,4,\cdots\right\}$. Define the $\displaystyle P_n$ as being the set of $\displaystyle n$ natural numbers in $\displaystyle [1,b]$. It is clear that $\displaystyle \Delta x_i=1$. Now consider the $\displaystyle i$th point in the partition. This point will be $\displaystyle i$ (since we defined the partitions as the set of natruals). Then the interval $\displaystyle [x_{i},x_{i+1}]$ will be the interval $\displaystyle [i,i+1]$, so on any interval $\displaystyle M_i=\sup_{x_{i}\leqslant x\leqslant x_{i+1}}f(x)=\sup_{i\leqslant x\leqslant i+1}f(x)=f(i)$ since $\displaystyle f$ is monotonically decreasing. So $\displaystyle U\left(P_n,f\right)=\sum_{i=1}^{n-1}M_i\cdot\Delta x_i=\sum_{i=1}^{n-1}f(i)$. So on any interval $\displaystyle [1,b]$ we have that $\displaystyle \int_1^b f~dx\leqslant\sum_{i=1}^{n-1}f(i)$. So now letting $\displaystyle b\to\infty\implies P_n\to\mathbb{N}$ which in turn implies $\displaystyle n\to\infty$ (since there are infinitely many elements of $\displaystyle \mathbb{N}$) so $\displaystyle \int_1^{\infty} f~dx\leqslant\sum_{x=1}^{\infty}f(x)$ and since the RHS converges by the hypothesis this concludes the proof.

Part two: $\displaystyle \int_1^{\infty}f~dx\text{ converges}\implies\sum_{x=1}^{\infty}f(x)\text{ converges}$. Define $\displaystyle [1,b]$ and $\displaystyle P_n$ as before, consqequently we have that $\displaystyle \Delta x_i=1$ again. Except this time $\displaystyle m_i=\inf_{x_i\leqslant x\leqslant x_{i+1}}f(x)=\inf_{i\leqslant x\leqslant i+1}=f(i+1)$. So $\displaystyle L\left(P_n,f\right)=\sum_{i=1}^{n-1}m_i\cdot\Delta x_i=\sum_{i=1}^{n-1}f(i+1)$. Now as was stated $\displaystyle L\left(P_n,f\right)\leqslant \int_a^b f~dx$ so this implies $\displaystyle \sum_{i=1}^{n-1}f(i+1)\leqslant\int_1^b f~dx$. So once again letting $\displaystyle b\to\infty\implies P_n\to\mathbb{N}$ which in turn implies $\displaystyle n\to\infty$ so $\displaystyle \sum_{x=1}^{\infty}f(x+1)\leqslant \int_1^{\infty} f~dx$. Now since the LHS may be written as $\displaystyle \sum_{x=1}^{\infty}f(x)-f(1)$ and the RHS converges this concludes the proof.

Now combinging parts one and two gives $\displaystyle \sum_{x=1}^{\infty}f(x)\text{ converge}\Longleftrightarrow\int_1^{\infty}f~dx\te xt{ converges}\quad\blacksquare$

Hows that look?