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

You know how to evaluate improper integrals right?

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

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

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

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

ok let me explain why I am confused

@zepdrix can you help?

yes it does

I don't understand that can you explain it?

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

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

Thanks

ok that statement seems logical.

also differentiability implies continuity ...

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

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

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

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

sorry ... i was out

no problem

you just need to show f(n+1)

physics graduate student

with a bit experience on analysis

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

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

this does not converge

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

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

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

yes ... that it is

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

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

is calc II differential equatons or multivariable calc?

Honestly I always wondered what the difference is. lol

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

what are spivak's calculus?

Then we go on to a class called differential equations.

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

so what you call analysis is the classes below calculus?

So This is the hard part? The analysis?

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

no no it's okay ...

all right good luck.

Thanks again!!!!!