confirm the integral test can be applied to the series.

- chrisplusian

confirm the integral test can be applied to the series.

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- chrisplusian

\[\sum_{1}^{\infty}\frac{ n }{n^{4} +1 }\]

- anonymous

You know how to evaluate improper integrals right?

- chrisplusian

So I know that you have to show that it is positive, continuos, and decreasing for
\[x \ge 1\]

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## More answers

- anonymous

Well first you have to check the the series is decreasing and approaching 0. So if you aren't sure it is decreasing, find its derivative to show it is decreasing over x.

- chrisplusian

and that I have to show that \[A_{n}=f(n) \]

- anonymous

Its derivative would be (x^4 + 1) - ( 4x^4 ) which would give you a decreasing function in the denominator.

- chrisplusian

That is where I run into the problem..... taking the first derivative gives me:
\[\frac{ -3x^{4} +1 }{ (x^{4} +1)^2 }\]

- chrisplusian

@MarcLeclair did you use the quotient rule to get the derivative?

- anonymous

Yeah so it would decrease for all x when x is greater than 1

- anonymous

Yes i did I just wrote the top part :) because the denominator wont dictate if it is negative or not because it is squared

- chrisplusian

ok let me explain why I am confused

- chrisplusian

the only way I know to show that the function is decreasing is by taking the first derivative and then finding the critical values. Then I use a sign chart to determine if it is increasing or decreasing. The problem with this one is finding the critical values. Maybe my algebra is a little rusty.

- anonymous

Well this is weird, when I was doing it ( I just finished cal 2 here, took my final few weeks ago) I did it in my book and it only found the first derivative. It didn't go all the way showing its critical values, it just showed the numerator was decreasing over x when x was greater than a certain value. If you're doing it by finding those I guess I'm the wrong person to help :P

- chrisplusian

well the critical values of the first derivative are the places where the graph can change from increasing to decreasing so we are required in my class to show that there are either no critical values that are greater or equal to one, or that if there are they are continuosly decreasing.

- chrisplusian

@zepdrix can you help?

- chrisplusian

@experimentX ?

- experimentX

yes it does

- experimentX

for a series decreasing to zero,
\[ \int_a^b f(x)dx - f(a) \le \sum_{n=a}^b f(n) \le \int_a^b f(x)dx + f(a)\]

- chrisplusian

I don't understand that can you explain it?

- experimentX

this is too long ... i'll upload a snapshot

- chrisplusian

does that follow the epsilon delta proofs? We were told not to do the epsilon delta parts of these

- chrisplusian

Thanks

- experimentX

##### 1 Attachment

- chrisplusian

I will have to go print that at office max.... I have a macbook air and it is to big to show on this lillte screen. Hopefully that will help thank you.

- experimentX

integral test is stronger than ratio test or root test in general. for you case,
\[ \int_1^\infty \frac{x}{x^4 + 1 }dx < \int_1^\infty \frac{x}{x^4}dx\]
is less than inifinity

- chrisplusian

ok that statement seems logical.

- chrisplusian

The problem I am running into is that my professor wants me to show four steps before I can use the integral test and there will be problems on our test that say specifically to use the integral test, which in turn means to show these four things: 1)f(x) is positive for x ge 1. 2) f(x) is continuous for x ge 1. 3) f is decreasing for x ge 1. 4)\[a_{n}=f(n) \] for n= positive integer

- chrisplusian

The problem I ma running into over and over is that the functions I am given are very hard to find the critical values of the f prime function and thereby make it very difficult to show that the original function is decreasing for all x ge 1

- experimentX

f(x) satisfies all condition, the poles are all complex at |z| = 1, the function is decreasing to zero ... to show that take derivative, put x>0 ... lim x->infty f(x) = 0 is not hard to show.

- experimentX

also differentiability implies continuity ...

- experimentX

to show positive, assume, f(x)>0, it immediately gives you that for all x>0, the f(x) is positive since x^4 + 1 is never negative.

- chrisplusian

what do you mean "f(x) satisfies all condition, the poles are all complex at |z| = 1"?

- chrisplusian

I showed that it was positive the same way you did. assuming that f(x)>0 and proving it that way.

- chrisplusian

also why do you say that the function is decreasing to zero?

- chrisplusian

I am sorry to ask so many questions, just having a difficult time with it

- experimentX

sorry ... i was out

- chrisplusian

no problem

- experimentX

you just need to show f(n+1)

- chrisplusian

oh that makes sense..... but fairly long in this problem, yet that makes sense. its alost like you are showing it is monotonic for that interval.

- experimentX

there is a thing called singularity ... when the function becomes infinite, for that x^4 +1 = 0 => all roots are complex. there is singularity on real line.

- chrisplusian

I have never heard that before. I wish I knew math the way you do. Are you a graduate student or professor?

- experimentX

physics graduate student

- experimentX

with a bit experience on analysis

- chrisplusian

thats awesome, I am working on my B.S. majoring in engineering minor in physics.

- chrisplusian

Do you have to prove the above mentioned statements for ALL x ge 1 or can it be for a different bound ? ex......

- experimentX

yes ... those are required condition for integral test. for eg, integral test don't work on alternating series.

- chrisplusian

\[\sum_{1}^{\infty}\frac{ \ln(n) }{ \sqrt{n} }\]

- experimentX

this does not converge

- chrisplusian

In this case it is increasing for \[0 \le x \le e^2\]

- chrisplusian

Does the fact that it is increasing automatically show that you cannot use the integral test?

- experimentX

you know that you can always partition the series ... consider two separate cases. 1) n < e^2 and n>e^2
for finite sum always converges given that there is no singularity. for infinite value, use integral test.

- chrisplusian

Some of the questions we have are strictly to decide if the integral test can be used to show convergence or divergence

- chrisplusian

So as long as I use the lower bound of integration as 8 then I can use the integral test?

- experimentX

yes ... that it is

- chrisplusian

Ok can I ask you an unrelated question? It is about physics with calc one and two?

- experimentX

lol ... calc is heavy sub on physics. calc I and II are for undergrad juniors. I had mathematical physics which focuses on many areas of math mostly non linear DE, PDE, Complex analysis, Tensors, Linear algebra, Statistics, Numerical analysis

- experimentX

my courses assumes that I have enough experience with calc I, II and III

- chrisplusian

I am in a small school and have completed physics with calc one and physics with calc two. I was grouped in with the regular physics students because of the class size (only three for physics with calc one, and two students for physics with calc two). Our professor never taught a single thing in regards to calculus based physics. My textbook did not correspond to what he was teaching because of the calculus element. I have to go on and take thermal dynamics, statics, etc. Plus I am going to have to take 18 hours in physics to get the minor in physics. I am contemplating filing a grievance and asking to be taught the material with the calculus element because the school did not meet the requirements of the statewide numbering system (Florida). But my professor is saying not to worry about it because it is only important that I understand the concepts, that the calc is not that important. What do you think? Is it a mistake to proceed without knowing the calculus side?

- chrisplusian

I am transferring to University of Florida in five months, and I saw some of their phyisics with calculus questions from their study review....... I had no clue.

- experimentX

so far i know calc I is elementay calculus, calc II is differential equations and calc III is multivariable calc ... these are basic courses commonly taught for all physics/engg student. you should have mastery of it. I recommend it. it seems that you are currently doing analysis. this is a bit hard part.

- experimentX

so far i have seen, this is probably with linear algebra this is usually taught to freshmen or sophomore. of course math gets messy on III and IV years, concept is important and so it problem solving.

- chrisplusian

I have taken calculus one, this is the end of calculus two, and I have already completed physics with calculus one and physics with calculus two at my college physics with calculus two is just a co-requisite of physics with calculus two

- experimentX

is calc II differential equatons or multivariable calc?

- chrisplusian

Honestly I always wondered what the difference is. lol

- experimentX

what did you study then?
for first two years at uni
http://tutorial.math.lamar.edu/ <-- this is more than enough

- chrisplusian

here they use a numbering system and to complete a class you have to complete certain objectives. In this class CALC TWO you have to go through all the way to the ratio and root test. First part of the semester we learned the

- experimentX

lol ... this is not calc, it's a part of analysis. http://en.wikipedia.org/wiki/Real_analysis
when calc get's worse, usually analysis comes to rescue.

- experimentX

although i've heard that if you study spivak's calculus, then things are messy

- chrisplusian

what are spivak's calculus?

- chrisplusian

Cacl three topics here are:
polar equations
vectors and geometry of space
vector-valued functions
functions of several variables
multiple integration
and vector analysis

- chrisplusian

Then we go on to a class called differential equations.

- experimentX

that's calculus book http://www.amazon.com/Calculus-4th-Michael-Spivak/dp/0914098918

- chrisplusian

so what you call analysis is the classes below calculus?

- experimentX

it's hard part ... more mathematical than usual mathematics you do in physics or engineering. close to pure math.

- chrisplusian

So This is the hard part? The analysis?

- experimentX

look at the topics http://www.amazon.com/Principles-Mathematical-Analysis-Third-Walter/dp/007054235X you will find that it is similar to calculus but not calculus exactly

- chrisplusian

Thanks you have been a great help. WIsh I could give you a few more medals lol.

- experimentX

no no it's okay ...

- experimentX

the thing is calculus is too smooth like integration, derivatives, etc ... with analysis you will understand the underlying theory behind it. take a look ... it's more rigorous than calc but don't spend much time in it.

- experimentX

one eg: if you have infinite nested roots like this one, it's easy to calculate it's value, it's easy part ... the hard part comes when you have to show that this value exists using some simple theorems. basically analysis deals with these stuff.
|dw:1369672079150:dw|

- experimentX

all right good luck.

- chrisplusian

Thanks again!!!!!

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