Integral test

Mathstud28 · #1 $Old$ December 24th, 2008, 11:21 PM

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:

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

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

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

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

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

Taking $\alpha(x)=x$

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

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

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

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

Ok now onto the question

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

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

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

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

Hows that look?

Mathstud28 · #2 $Old$ December 25th, 2008, 11:46 PM

I have a question pertaining to this. Before I did it as such I considered just partitioning $[1,\infty)$ by the naturals. Of course I realized that there were a plethora of problems with this but one in particular I wanted to ask about. I know that a partition $P$ of $[a,b]$ must have finite cardinality but does this still apply when either/or $a,b$ is $\infty,-\infty$ ? I think the answer is yes, since technically $\int_a^{\infty} f~d\alpha$ does not make sense in the Riemann-Stieltjes sense (we of course obviate this by writing $\lim_{b\to\infty}\int_a^b f~d\alpha$ .