# Sum of limit

## Homework Statement

Given the function in x
$$f_n(x)=sin^nx (n=1,2,3,...)$$
For this ##f_n(x)##, consider the definite intergral
$$I_n=\int_{0}^{\pi/2}f_n(x)sin2xdx$$
a) Find ##I_n##
b) Hence the obtain
$$lim_{n→∞}(I_{n-1}+I_n+I_{n+1}+...+I_{2n-2})=\int_0^W\frac{X}{Y+x}dx$$
Find X,Y,Z.

## Homework Equations

b) I think we may use Riemann sum, and the answer in a, help we in b, but I can't do it.

## The Attempt at a Solution

a) ##I_n=\int_{0}^{\pi/2}f_n(x)sin2xdx=2\int_{0}^{\pi/2}sin^{n+1}cosxdx=2\int_{0}^{\pi/2}sin^{n+1}d(sinx)##
$$I_n=\frac{2}{n+2}$$
b) Dont have any attempt.

I need some books about the calculus I that have some problems like this
Thank you very much

Last edited:

haruspex
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Have you tried swapping the order of the steps, i.e. performing the sum before integrating?

Hamal_Arietis
It means:$$lim_{n→∞}\Sigma=lim_{n→∞}2(\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{2n})(1)$$
The Riemann sum must have form: $$\int = lim_{n→∞}\Sigma _{i=0}^n f(i)\Delta x$$
I can't make (1) to appear element ##\Delta x##.

haruspex
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Can you use the harmonic sum bounds given by the log function and Euler's constant?
Trouble is, it does not give you an integral, but the actual answer.

fresh_42
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Can you use the harmonic sum bounds given by the log function and Euler's constant?
Trouble is, it does not give you an integral, but the actual answer.
But your previous answer has been a good idea (it's been what I was looking for), to integrate the sum of sines which has a closed form. With the substitution of the OP (##u = \sin x##) one gets an integral of one quotient of polynomials in ##u##.

haruspex
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But your previous answer has been a good idea (it's been what I was looking for), to integrate the sum of sines which has a closed form. With the substitution of the OP (##u = \sin x##) one gets an integral of a quotient of polynomials in ##u##.
Yes, but I could not see how to massage it into the required form. Maybe I missed something.

fresh_42
Mentor
Yes, but I could not see how to massage it into the required form. Maybe I missed something.
I haven't seen it either, but as long as nobody requires ##X## and ##Y## to be constants ...

haruspex
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I haven't seen it either, but as long as nobody requires ##X## and ##Y## to be constants ...
I would assume they have to be.

The answer W=1; X=2; Y=1 and the integral equals ln4. I think it may not use Euler's constant because It is the question for high school students. May be difficult but they dont learn about harmonic sum and Euler's constant.

haruspex
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The answer W=1; X=2; Y=1 and the integral equals ln4. I think it may not use Euler's constant because It is the question for high school students. May be difficult but they dont learn about harmonic sum and Euler's constant.
Well, you don't need Euler's constant as such. You just need the bounded form of the Riemann sum: with suitable facts about f, ##\Sigma_m^{n-1}f(r)<=\int_m^nf <= \Sigma_{m+1}^n f(r)##, or something like that.

fresh_42
Mentor
One could use the Euler-Mascheroni constant and write the sum ##I_{n-1}+\ldots +I_{2n-2} = I_{-1}+\ldots +I_{2n-2} - I_{-1} - \ldots - I_{n-2}##.

Hamal_Arietis
haruspex
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One could use the Euler-Mascheroni constant and write the sum ##I_{n-1}+\ldots +I_{2n-2} = I_{-1}+\ldots +I_{2n-2} - I_{-1} - \ldots - I_{n-2}##.
Yes, but Hamal wants to avoid appealing to that as a standard result. As I posted, you can effectively achieve the same using Riemann sums, which Hamal wanted to do in the first place.

One could use the Euler-Mascheroni constant and write the sum ##I_{n-1}+\ldots +I_{2n-2} = I_{-1}+\ldots +I_{2n-2} - I_{-1} - \ldots - I_{n-2}##.
If I use Euler-Mascheroni constant, how to make ##lim_{n→∞}∑=∫##?
Yes, but Hamal wants to avoid appealing to that as a standard result. As I posted, you can effectively achieve the same using Riemann sums, which Hamal wanted to do in the first place.
If that can make answer, can you show me? The Riemann sum is just my idea. Maybe have another method.

In part A of problem, they give some question, And that is anwer I solved, I think maybe help you.
Given the funtion of x $$f_n(x)=sin^nx(n=1,2,3...)$$
Consider the cases in which the equality
$$lim_{x→0}\frac{a-x^2-(b-x^2)^2}{f_n(x)}=c$$
holds for three real numbers a,b,c.
a) We have ##a=b^{O}##. Find O
b) When n=2, if c=6. Find b
c) When n=4 find b and c.
My solution:
a) Because c is real number so
$$lim_{x→0}\frac{a-x^2-(b-x^2)^2}{f_n(x)}=c⇔lim_{x→0}(a-x^2-(b-x^2)^2)=c.lim_{x→0}f_n(x)=0⇒a=b^2$$
b) $$lim_{x→0}\frac{a-x^2-(b-x^2)^2}{f_n(x)}=lim_{x→0}\frac{-x^2(x^2+2b-1)}{sin^2x}$$
$$lim_{x→0}\frac{-x^2(x^2+2b-1)}{sin^2x}=6⇔b=\frac{7}{2}$$
c) b=2 ;c=-1

haruspex
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In part A of problem, they give some question, And that is anwer I solved, I think maybe help you.
Given the funtion of x $$f_n(x)=sin^nx(n=1,2,3...)$$
Consider the cases in which the equality
$$lim_{x→0}\frac{a-x^2-(b-x^2)^2}{f_n(x)}=c$$
holds for three real numbers a,b,c.
a) We have ##a=b^{O}##. Find O
b) When n=2, if c=6. Find b
c) When n=4 find b and c.
My solution:
a) Because c is real number so
$$lim_{x→0}\frac{a-x^2-(b-x^2)^2}{f_n(x)}=c⇔lim_{x→0}(a-x^2-(b-x^2)^2)=c.lim_{x→0}f_n(x)=0⇒a=b^2$$
b) $$lim_{x→0}\frac{a-x^2-(b-x^2)^2}{f_n(x)}=lim_{x→0}\frac{-x^2(x^2+2b-1)}{sin^2x}$$
$$lim_{x→0}\frac{-x^2(x^2+2b-1)}{sin^2x}=6⇔b=\frac{7}{2}$$
c) b=2 ;c=-1
I urge you to try what I proposed in post #10. Draw a diagram to illustrate the inequalities.

Hamal_Arietis
I solved this problem. Firstly, I see
##I_{n-1}=\frac{2}{n+1};I_n=\frac{2}{n+2};...;I_{2n-2}=\frac{2}{n+n}##
So let $$P_i=\frac{2}{n+i}=\frac{1}{n}\frac{2}{1+\dfrac{i}{n}}=\Delta x .f(x_i)$$
If ##i=1;n→∞⇒x→0## and ##i=n;n→∞;x=1##
So we have the Riemann sum:
$$lim_{n→∞}(I_{n-1}+I_n+I_{n+2}+...)=lim_{n→∞}\Sigma_{i=1}^{n}\frac{1}{n}\frac{2}{1+\dfrac{i}{n}}=∫_{0}^{1}\frac{2dx}{x+1}$$
Thanks for all helping

haruspex
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I solved this problem. Firstly, I see
##I_{n-1}=\frac{2}{n+1};I_n=\frac{2}{n+2};...;I_{2n-2}=\frac{2}{n+n}##
So let $$P_i=\frac{2}{n+i}=\frac{1}{n}\frac{2}{1+\dfrac{i}{n}}=\Delta x .f(x_i)$$
If ##i=1;n→∞⇒x→0## and ##i=n;n→∞;x=1##
So we have the Riemann sum:
$$lim_{n→∞}(I_{n-1}+I_n+I_{n+2}+...)=lim_{n→∞}\Sigma_{i=1}^{n}\frac{1}{n}\frac{2}{1+\dfrac{i}{n}}=∫_{0}^{1}\frac{2dx}{x+1}$$
Thanks for all helping
Neat. Well done.