Question

Integrals and Intervals of Convergence

Answer 1

@ganeshie8 @Hero @campbell_st

Answer 2

whats the question

Answer 3

How do I set this up. Am I supposed to make a series expansion of this? Becuase that's what the lesson is. And they threw in the interval of convergence in there, even though that's the next lesson (But I figure it's the interval in which the series can convergence). But yes, i need to set this up since it said to find it exactly and all I've been doing approximating...

Answer 4

look you have attached the wrong pic, it has nothing to do with interval of convergence

Answer 5

I see only one pic

Answer 6

I think they just mentioned it there in the question....I posted the wrong picture first, is that what you're looking at? I posted a new one

Answer 7

Do they first define the function f(x)= summation ?

Answer 8

I dont see the word "interval" in the picture

Answer 9

f(x) = summation from n equals 1 to infinity of the product of the quotient of the quantity n plus 1 and 5 to the n plus 1 power and x to the nth power with an INTERVAL of convergence, –5 < x < 5. Find exactly the value of the integral from 0 to 4 of of x, dx. Your answer will be a positive integer. Type your answer in the space below (ex. 2).

Numerical Answers Expected!

Answer 10

That's the text of the question, and they spelled out the summation sorry.

Answer 11

It says Question 5 in the corner correct? If so that's the question I'm talking about

Answer 12

are you on mobile ?

Answer 13

No, here it is again

Answer 14

ohk..

Answer 15

You may change the order of `summation` and `integral` for a power series :
\[\begin{align}\int\limits_0^4f(x)&=\int\limits_0^4\,dx\sum\limits_{n=1}^{\infty}\frac{n+1}{5^{n+1}}x^n\,dx\\~\\
&=\sum\limits_{n=1}^{\infty}\int\limits_0^4\frac{n+1}{5^{n+1}}x^n\,dx\\~\\
&=\sum\limits_{n=1}^{\infty}\frac{n+1}{5^{n+1}}\int\limits_0^4x^n\,dx\\~\\
&=\cdots
\end{align}\]

Answer 16

Ok, how do I find the summation and I figure that wil cancel out with the integral to get me the positive integer, correct? I'm referring to the ns when I say cancel out

Answer 17

first evaluate the integral

Answer 18

Ok \[\Large \frac{ 4^{n+1} }{ n+1 }\]

Answer 19

\[\begin{align}\int\limits_0^4f(x)&=\int\limits_0^4\,dx\sum\limits_{n=1}^{\infty}\frac{n+1}{5^{n+1}}x^n\,dx\\~\\
&=\sum\limits_{n=1}^{\infty}\int\limits_0^4\frac{n+1}{5^{n+1}}x^n\,dx\\~\\
&=\sum\limits_{n=1}^{\infty}\frac{n+1}{5^{n+1}}\int\limits_0^4x^n\,dx\\~\\
&=\sum\limits_{n=1}^{\infty}\frac{n+1}{5^{n+1}}\frac{4^{n+1}}{n+1}\\~\\
&=\sum\limits_{n=1}^{\infty}\left(\frac{4}{5}\right)^{n+1}\\~\\
&=\cdots
\end{align}\]

Answer 20

that just well known geometric series right

Answer 21

Oh! That's cool. I got it from here

Answer 22

looks that converges to \(4\)
http://www.wolframalpha.com/input/?i=%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7D+%284%2F5%29%5E%28n%2B1%29