I tried to approach integrating by filling the space with infinitely many infinitely narrow rectangles.

$a$ is the left bound

$b$ is the right bound

$n$ means the number of rectangles, approaches $\infty$

$k$ means the k-th rectangle

My integral is described by the following formula:

$$ \lim_{n \to \infty} \sum_{k=1}^{n} \frac{b-a}{n} \cdot f(a + \frac{k(b-a)}{n}) $$

The rough idea in the picture below:

enter image description here

Let's solve for $f(x) = \frac{1}{x}$, a = 1, b = 3

$$ \lim_{n \to \infty} \sum_{k=1}^{n} \frac{2}{n} \cdot \frac{1}{1 + \frac{2k}{n}} = \lim_{n \to \infty} \left(\frac{2}{n+2}+\frac{2}{n+4}+\frac{2}{n+6}+ \dots\right) $$ The sum clearly diverges, but I don't know why.. any ideas? Is this way of integrating possible? If not, why? Are my formulas correct?


Your sum should only go up to $n$. You have otherwise properly described the right hand Riemann sum. The sum has a finite number of terms, so it does not diverge. You take the limit after you do the sum and should get $\log 3$.

  • $\begingroup$ $n$ approaches $\infty$ though? That's why I put it above the sum. What else should I limit $k$ to? $\endgroup$ – Mateusz Sowiński Feb 1 '19 at 19:06
  • $\begingroup$ That is the point of the limit out front. You evaluate the sum for a given $n$, then take the limit. The sum is bounded by $2$, so the limit cannot diverge either $\endgroup$ – Ross Millikan Feb 1 '19 at 19:11
  • $\begingroup$ @MateuszSowiński $n$ approaches $\infty$, but we're dealing with a sequence of sums—with the $n$th sum having $n$ as its upper limit. You've got $k$ approaching $\infty$ in each one. $\endgroup$ – timtfj Feb 1 '19 at 19:11
  • $\begingroup$ @RossMillikan I would do that if I wanted an approximation, right? I want an exact number though, that's why I'm adding an infinite amount of these rectangles $\endgroup$ – Mateusz Sowiński Feb 1 '19 at 19:16
  • $\begingroup$ @MateuszSowiński Increasing $n$ = using more rectangles for the next sum. Increasing $k$ = counting through the $n$ rectangles for a given $n$, so $k$ only ever goes up to $n$. $\endgroup$ – timtfj Feb 1 '19 at 19:18

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