# Summation of series using definite integral

I learned that definite integral gives the signed area under a curve by dividing the curve into small rectangular strips and "making" its width shrink to zero.

Using this knowledge summation of certain series can be found. I converted definite integral in this form

$$\int_{a}^{b} f(x) \, \mathrm{d}x = \lim_{n\to\infty} \frac{b-a}{n} \sum_{r=1}^{n} f\left( a + \left(\frac{b-a}{n}\right)r \right)$$

However, another equivalent form is known to exist, provided below, and I find it much more convenient to use that form. I tried to manipulate the sum to arrive at the equivalent form but so far I've not able to succeed.

$$\int_{a}^{b} f(x) \, \mathrm{d}x = \lim_{n\to\infty} \frac{1}{n} \sum_{r=g(n)}^{h(n)} f\left(\frac{r}{n}\right), \quad \text{where } \begin{cases} \lim_{n\to\infty} \frac{g(n)}{n} = a, \\ \lim_{n\to\infty} \frac{h(n)}{n} = b. \end{cases}$$

Can someone help (give me a hint perhaps) on how to convert to the second much more convenient expression.

I find second expression much more convenient because the limits of integration can be very easily found using it.

Any help will be appreciated.

EDIT: I've still not able to derive the equivalent relation mentioned above. Can someone hint me?

• Hmmm.. why isn't someone responding to my question? Sorry if it sounds a bit silly question. Feb 4, 2019 at 9:11
• The notation is a bit unclear to me. For the second sum to make sense, it should hold that $g(n)$ and $h(n)$ are always integers with $g(n) \leq h(n)$ for any $n \in \mathbb N$. Mar 5, 2019 at 21:00
• Could you give a practical example for the second expression?
– user
Mar 5, 2019 at 21:07
• @user Consider when lower index of summation is 1 and upper index is 2n. Limits on integration will be given by Mar 5, 2019 at 21:10
• ...a= 1/n (n tends to infinity) = 0 and b = 2n/n (n tends to infinity) = 2 Mar 5, 2019 at 21:11

It more make sense that we focus on the behavior of $$\frac1n\sum_{r=h(n)}^{g(n)}f(r/n)$$ as $$n\to \infty$$. Think of the $$1/n$$ as the width of each subinterval of $$[a,b]$$, and think of $$r/n$$ as $$x$$-values in each subinterval. As $$n\to\infty$$, each subinterval approaches a single point, namely $$\lim_{n\to\infty}r/n$$. So it's just a sum of $$\text{height}\times\text{width}$$.