Integral as approximation to summation

[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

 

[PeroK]

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?

 

[songoku]

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

 

[PeroK]

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.

 

[Office_Shredder]

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?

 

[songoku]

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...

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

 

[fresh_42]

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.

 

[Office_Shredder]

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.

 

[songoku]

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

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

 

[Office_Shredder]

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?

 

[Delta2]

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.

 

[fresh_42]

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.

 

[sysprog]

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

 

[fresh_42]

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.

 

[songoku]

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 $$\int_0^1(1+x)^{-1}dx$$.

No I am not. I have googled it and read a little bit about it, it is really helpful

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?

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

 

[Office_Shredder]

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.

 

[songoku]

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