Using the Integral Test to Show Sum is Less Than pi/2

[Kqwert]

Homework Statement

Use the integral test to show that the sum of the series

gif.latex

$\sum_{n=1}^\infty \dfrac{1}{1+n^2}$

is smaller than pi/2.

Homework Equations

The Attempt at a Solution

I know that the series converges, and the integral converges to pi/4. As far as I´ve understood, this does however not mean that the series converges to pi/4. How should I solve this?

 

[fresh_42]

The integral criterion comes along with an estimation, so what is your version you will be allowed (expected) to use?

 

[Ray Vickson]

Homework Statement

Use the integral test to show that the sum of the series

gif.latex

$\sum_{n=1}^\infty \dfrac{1}{1+n^2}$

is smaller than pi/2.

Homework Equations

The Attempt at a Solution

I know that the series converges, and the integral converges to pi/4. As far as I´ve understood, this does however not mean that the series converges to pi/4. How should I solve this?

Draw a picture of $y = 1/(1+x^2)$ and superimpose a picture of the step-function $y = 1/(1+n^2), n=1,2,3, \ldots.$

 

[Kqwert]

The integral criterion comes along with an estimation, so what is your version you will be allowed (expected) to use?

Not exactly sure what you are referring to here?

 

[Kqwert]

Draw a picture of $y = 1/(1+x^2)$ and superimpose a picture of the step-function $y = 1/(1+n^2), n=1,2,3, \ldots.$

Do you mean plotting the function y(x) in the same graph as the rectangles from the step-function? And then noticing that the rectangles from the step-function is an underestimation with respect to the function y(x)?

 

[fresh_42]

I was referring to the integral test. Wikipedia has $\int \ldots \leq \sum \ldots \leq \int \ldots$ and my language version $\sum \ldots \leq \int \leq \ldots \sum \ldots$ so there is an estimation for all possible cases. They probably come out of the proof for the criterion. If you do not want to use these ready made estimations, use Ray's idea in post #3.

 

[fresh_42]

Do you mean plotting the function y(x) in the same graph as the rectangles from the step-function? And then noticing that the rectangles from the step-function is an underestimation with respect to the function y(x)?

Use an under- and an overestimation, depending on how you count: starting with n=0 or n=1 for the step functions.

 

[Kqwert]

Use an under- and an overestimation, depending on how you count: starting with n=0 or n=1 for the step functions.

Not exactly sure how do this.. do you know of any relevant examples anywhere?

 

[fresh_42]

Not exactly sure how do this.. do you know of any relevant examples anywhere?

Draw the function $f(x)=y=\frac{1}{1+x^2}$ in an $(x,y)-$ coordinate system. Then draw the functions
$$
f_+(x)= \frac{1}{1+\lfloor x\rfloor^2} \text{ if } \lfloor x \rfloor \leq x \leq 1+ \lfloor x \rfloor \text{ and } f_-(x)= \frac{1}{1+(\lfloor x\rfloor + 1)^2} \text{ if } \lfloor x \rfloor \leq x \leq 1 + \lfloor x \rfloor
$$
and compute $\sum_n f_+(n)\; , \;\int f(x)\,dx=\int \frac{1}{1+x^2}\,dx\; , \;\sum_n f_-(n)$ in appropriate limits.

If in doubt, draw it out!

 

[Kqwert]

Draw the function $f(x)=y=\frac{1}{1+x^2}$ in an $(x,y)-$ coordinate system. Then draw the functions
$$
f_+(x)= \frac{1}{1+\lfloor x\rfloor^2} \text{ if } \lfloor x \rfloor \leq x \leq 1+ \lfloor x \rfloor \text{ and } f_-(x)= \frac{1}{1+(\lfloor x\rfloor + 1)^2} \text{ if } \lfloor x \rfloor \leq x \leq 1 + \lfloor x \rfloor
$$
and compute $\sum_n f_+(n)\; , \;\int f(x)\,dx=\int \frac{1}{1+x^2}\,dx\; , \;\sum_n f_-(n)$ in appropriate limits.

If in doubt, draw it out!

Thank you very much! However, is that the easiest way of doing it? This is Calculus 1, so I was thinking there was a more simple approach? If not, sorry.

 

[fresh_42]

Thank you very much! However, is that the easiest way of doing it? This is Calculus 1, so I was thinking there was a more simple approach? If not, sorry.

Yes, using Wikipedia is easier. There you can find the estimations which are needed.

You can also look for more appropriate upper / lower sums, but I think $\sum_n \frac{1}{n^2}$ which it is in the end is already easy, because you have $\frac{\pi^2}{6} \pm c$ with a constant $c$ as results.

Or you look in your proof: it should contain an upper and lower bound. But if you look at the graph I mentioned, it is easier than the formulas suggest. You don't have to draw an exact graph, some scribbling will do.

 

[Kqwert]

Yes, using Wikipedia is easier. There you can find the estimations which are needed.

You can also look for more appropriate upper / lower sums, but I think $\sum_n \frac{1}{n^2}$ which it is in the end is already easy, because you have $\frac{\pi^2}{6} \pm c$ with a constant $c$ as results.

Or you look in your proof: it should contain an upper and lower bound. But if you look at the graph I mentioned, it is easier than the formulas suggest. You don't have to draw an exact graph, some scribbling will do.

Like under "Remark" on the wiki page? https://en.wikipedia.org/wiki/Integral_test_for_convergence
I.e. I can calculate the lower and upper bound of my series quite easily.

 

[fresh_42]

Yes, and that is basically what the functions $f_{\pm}(x)$ do as well. As they are step functions with the step length $1$, their integrals equal the sums and the function $f(x)$ (and thus also its integral) is in between.

And if you change the Wikipedia page to "German" you get directly the formula you need.

 

[Ray Vickson]

Not exactly sure how do this.. do you know of any relevant examples anywhere?

If $f(x) = 1/(1+x^2)$ and $t(n) = 1/(1+n^2)$, then when you plot $y = f(x)$ and the $y= t(n)$, you have that $t(n)$ is the area under a rectangular block with base $(n-1) \to n$ (width 1) and height $t(n)$. Since $t(x)$ is a strictly decreasing function we have
$$ t(n) < \int_{n-1}^n f(x) \, dx, \; n = 1, 2, 3, \ldots,$$
so $t(1)$ is less than the area under $y = f(x)$ from 0 to 1, $t(2)$ is less than the area from 1 to 2, etc.

If you want, you can also get a lower bound on the sum by noting that
$$t(n) > \int_{n}^{n+1} f(x) \, dx,$$
so $t(1)$ exceeds the area from 1 to 2, $t(2)$ exceeds the area from 2 to 3, etc.