# Integral as approximation to summation

• songoku
In summary: Once you understand how to get the sum from the integral, you just have to work backwards. Doing this a couple times will help you build intuition about what to look for.I think the OP isn't aware of the definition of an integral as the limit of a Riemann sum. It is almost straightforward that the sum $$\sum_{r=1}^n\frac{1}{n}(1+\frac{r}{n})^{-1}$$ is the Riemann sum of the integral $$\int_0^1(1+x)^{-1}dx$$.But if you are not allowed to use the definition of an integral as the limit
songoku
Homework Statement
Show that:
$$\sum_{r=1}^n \frac{1}{n} \left(1+\frac{r}{n} \right)^{-1}\approx \int_0^1 (1+x)^{-1} dx$$
Relevant Equations
Not sure
Writing down several terms of the summation and then doing some simplifying, I get:

$$\sum_{r=1}^n \frac{1}{n} \left(1+\frac{r}{n} \right)^{-1}= \frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...\frac{1}{2n}$$

How to change this into integral form? Thanks

songoku said:
Homework Statement:: Show that:
$$\sum_{r=1}^n \frac{1}{n} \left(1+\frac{r}{n} \right)^{-1}\approx \int_0^1 (1+x)^{-1} dx$$
Relevant Equations:: Not sure

Writing down several terms of the summation and then doing some simplifying, I get:

$$\sum_{r=1}^n \frac{1}{n} \left(1+\frac{r}{n} \right)^{-1}= \frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...\frac{1}{2n}$$

How to change this into integral form? Thanks
Have you tried drawing a graph? How could you simply estimate the area under a curve - any ideas?

etotheipi
PeroK said:
Have you tried drawing a graph?
No, I haven't. I don't know if I need to use graph. So I need to draw graph of ##y=\frac{1}{x+1}## ?

How could you simply estimate the area under a curve - any ideas?
Not sure. I have drawn graph of ##y=\frac{1}{x+1}## and maybe the area under the curve from x = 0 to x = 1 can be estimated by using area of trapezium?

Thanks

songoku said:
No, I haven't. I don't know if I need to use graph. So I need to draw graph of ##y=\frac{1}{x+1}## ?Not sure. I have drawn graph of ##y=\frac{1}{x+1}## and maybe the area under the curve from x = 0 to x = 1 can be estimated by using area of trapezium?

Thanks
A trapezium is good. But, maybe you don't even need to be that clever. Take a look the terms in the sum.

etotheipi
I would start with the integral, how would you try to estimate the area by drawing rectangles under the graph?

Then compare that to your sum.

An alternate way is based on looking at the sum, what do you think the ##\Delta x## is in the rectangles that were used to generate it?

PeroK said:
A trapezium is good. But, maybe you don't even need to be that clever. Take a look the terms in the sum.
I have looked at the terms in the sum for a while and nothing change...

Office_Shredder said:
An alternate way is based on looking at the sum, what do you think the ##\Delta x## is in the rectangles that were used to generate it?
So I need to use rectangles to estimate the area under the curve and ##\Delta x## will be the width of rectangle?

Thanks

It is important to tell us what you are allowed to use. E.g. if you know
$$H(n)=\log(n)+\gamma+O(1/n)$$
then it is immediately clear.

songoku said:
I have looked at the terms in the sum for a while and nothing change...So I need to use rectangles to estimate the area under the curve and ##\Delta x## will be the width of rectangle?

Thanks

Yes, if you do this then you can get the summation on the left without much issue.

fresh_42 said:
It is important to tell us what you are allowed to use. E.g. if you know
$$H(n)=\log(n)+\gamma+O(1/n)$$
then it is immediately clear.
I haven't learned about that equation. The scope is high school level of integration and summation properties

Office_Shredder said:
Yes, if you do this then you can get the summation on the left without much issue.
I think I can get the summation from the integration. But if I want to get the integral form from summation, how to proceed from what I did previously?

$$\sum_{r=1}^n \frac{1}{n} \left(1+\frac{r}{n}\right)^{-1}=\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}$$

or maybe I don't really need to simplify it, just leave it like this:
$$\sum_{r=1}^n \frac{1}{n} \left(1+\frac{r}{n}\right)^{-1}=\frac{1}{n} \left(\frac{n}{n+1}+\frac{n}{n+2}+...\frac{n}{n+n}\right)$$
$$=\frac{1}{n} \left( \frac{1}{1+\frac{1}{n}}+\frac{1}{1+\frac{2}{n}}+...+\frac{1}{1+\frac{n}{n}}\right)$$

I feel it will just take a little bit more work to get the integral form but I just don't know how to continue.

Thanks

Once you understand how to get the sum from the integral, you just have to work backwards. Doing this a couple times will help you build intuition about what to look for.

To take your last formula, you can write that as
$$\frac{1}{n} f(1/n) + \frac{1}{n} f(2/n)+...$$

Where ##f(x)=\frac{1}{1+x}##. Does that suggest anything?

I think the OP isn't aware of the definition of an integral as the limit of a Riemann sum. It is almost straightforward that the sum $$\sum_{r=1}^n\frac{1}{n}(1+\frac{r}{n})^{-1}$$ is the Riemann sum of the integral $$\int_0^1(1+x)^{-1}dx$$.

But if you are not allowed to use the definition of an integral as the limit of a Riemann sum (because you might not have been taught of it as well) then perhaps @fresh_42 suggestion but you said you don't know that equation either.

songoku said:
I haven't learned about that equation. The scope is high school level of integration and summation properties
That was primarily meant as a polite hint that the point
Relevant Equations:
is actually very important. It defines what we may use. It's not just you who didn't write anything there, it is unfortunately very common to write "none" or similar. As a consequence dialogues last often much longer than they should because writers and readers use different toolboxes.

Last edited:
sysprog
Am I being too much of a curmudgeon if I suggest that maybe the template should say 'relevant 'equations or inequalities''?

Delta2
sysprog said:
Am I being too much of a curmudgeon if I suggest that maybe the template should say 'relevant 'equations or inequalities''?
Without going to get too much off topic, I think, that wouldn't change much for various reasons. I would call it actually: "background to be used" and would emphasize it a lot more than "own effort". To me it is already an effort to describe the circumstances accurately.

sysprog
Delta2 said:
I think the OP isn't aware of the definition of an integral as the limit of a Riemann sum. It is almost straightforward that the sum $$\sum_{r=1}^n\frac{1}{n}(1+\frac{r}{n})^{-1}$$ is the Riemann sum of the integral $$\int_0^1(1+x)^{-1}dx$$.

Office_Shredder said:
Once you understand how to get the sum from the integral, you just have to work backwards. Doing this a couple times will help you build intuition about what to look for.

To take your last formula, you can write that as
$$\frac{1}{n} f(1/n) + \frac{1}{n} f(2/n)+...$$

Where ##f(x)=\frac{1}{1+x}##. Does that suggest anything?
By definition, definite integral is the limit of the Riemann sum :

$$\int_a^b f(x) dx=\lim_{n \to \infty} \sum_{i=1}^n \Delta x . f(x_i)$$Let the width of rectangle be ##\Delta x## and ##\Delta x=\frac 1 n##, and the height of rectangle be ##f(x)## where ##f(x)=\frac{1}{1+x}##

To determine the upper and lower limit of integration, can it be done by using algebra or I need to get it from graph?
fresh_42 said:
That was primarily meant as a polite hint that the point
Relevant Equations:
is actually very important. It defines what we may use. It's not just you who didn't write anything there, it is unfortunately very common to write "none" or similar. As a consequence dialogues last often much longer than they should because writers and readers use different toolboxes.
Oh, sorry, I did not realize that. Personally I don't think "relevant equations" part is not important but for this question I just don't know what the relevant equations are. What I did only plug in some numbers and simplify it using basic algebra, really have no idea what other formula I have to use.
Thank you for the reminder

songoku said:
Let the width of rectangle be ##\Delta x## and ##\Delta x=\frac 1 n##, and the height of rectangle be ##f(x)## where ##f(x)=\frac{1}{1+x}##

To determine the upper and lower limit of integration, can it be done by using algebra or I need to get it from graph?

The upper and lower limits of integration are the effective values of x you get at the start and end of your summation. For example if ##\Delta x = 1/n## and you sum up n terms, the width of your interval must be 1.

Office_Shredder said:
The upper and lower limits of integration are the effective values of x you get at the start and end of your summation. For example if ##\Delta x = 1/n## and you sum up n terms, the width of your interval must be 1.
I think I understand

Thank you very much for all the help and explanation Perok, Office_Shredder, fresh_42, Delta2

Delta2

## 1. What is the concept of "Integral as approximation to summation"?

The concept of "Integral as approximation to summation" is a mathematical technique used to estimate the sum of a series by calculating the area under a curve. It is based on the fundamental theorem of calculus, which states that the integral of a function can be used to find the area under the curve of that function.

## 2. How is the integral used to approximate a summation?

The integral is used to approximate a summation by first converting the series into a continuous function. This function is then integrated over a specific interval to find the area under the curve. This area represents an estimation of the sum of the series.

## 3. What is the advantage of using the integral as an approximation to summation?

The advantage of using the integral as an approximation to summation is that it allows for the estimation of sums that are difficult or impossible to calculate using traditional methods. It also provides a more precise estimation as the number of terms in the series increases.

## 4. Can the integral be used to approximate any type of series?

No, the integral can only be used to approximate series that can be represented by a continuous function. This means that the series must have a defined pattern and cannot have any gaps or jumps in the values.

## 5. How accurate is the integral as an approximation to summation?

The accuracy of the integral as an approximation to summation depends on the complexity of the series and the number of terms used in the calculation. Generally, the more terms used, the more accurate the estimation will be. However, there may be some error due to the limitations of the technique.

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