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Comparing summations with integrals

Problem Statement
1. Show that ##\sum_{n=1}^N \arctan{(n)} \geq N \arctan{(N)}-(1/2)\ln{(1+N^2)}##
2. Determine numbers ##\alpha##, ##\beta## such that ##\lim\limits_{N\to\infty} N^\alpha \sum_{n=1}^N \arctan{(n)} =\beta## with ##\beta \neq 0##
Relevant Equations
##\int_a^b f(x) dx= F(b)-F(a)##
1. ##\sum_{n=1}^N \arctan{(n)} \geq N \arctan{(N)}-(1/2)\ln{(1+N^2)} \iff \sum_{n=1}^N \arctan{(n)} \geq N \int_0^N \frac{1}{1+x^2} dx - \int_0^N \frac{x}{1+x^2} dx##

Where do I go from here? I've tried understanding this graphically, but to no avail.

2. Maybe this follows from finding an upper and lower bound for ##N^\alpha \sum_{n=1}^N \arctan{(n)}##, and then somehow applying the squeeze theorem?
 

fresh_42

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You should write the right hand side as an integral of ## \operatorname{arctan}## and the left as an upper Riemann sum for this integral.
 
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Problem Statement: 1. Show that ##\sum_{n=1}^N \arctan{(n)} \geq N \arctan{(N)}-(1/2)\ln{(1+N^2)}##
2. Determine numbers ##\alpha##, ##\beta## such that ##\lim\limits_{N\to\infty} N^\alpha \sum_{n=1}^N \arctan{(n)} =\beta## with ##\beta \neq 0##
Relevant Equations: ##\int_a^b f(x) dx= F(b)-F(a)##

1. ##\sum_{n=1}^N \arctan{(n)} \geq N \arctan{(N)}-(1/2)\ln{(1+N^2)} \iff \sum_{n=1}^N \arctan{(n)} \geq N \int_0^N \frac{1}{1+x^2} dx - \int_0^N \frac{x}{1+x^2} dx##

Where do I go from here? I've tried understanding this graphically, but to no avail.
This looks to me like a Riemann sum for an integral: ##\int_{x = 0}^N \arctan(x)~dx##, or maybe ##\int_{x = 1}^{N + 1} \arctan(x)~dx##. That's the approach I would take for starters.
schniefen said:
2. Maybe this follows from finding an upper and lower bound for ##N^\alpha \sum_{n=1}^N \arctan{(n)}##, and then somehow applying the squeeze theorem?
 
If we have ##\sum_{n=1}^N \arctan{(n)} \geq \int_0^N \arctan{(x)} dx##, then ##N^\alpha \sum_{n=1}^N \arctan{(n)}## is simply the latter multiplied by a constant ##N^\alpha##. To obtain this expression in the first place, would it be reasonable to find an upper bound to the upper Riemann sum? Does it exist?
 
By the way, out of curiosity, how does one get ##\int_0^N \arctan{(x)} dx## from ##N \int_0^N \frac{1}{1+x^2} dx - \int_0^N \frac{x}{1+x^2} dx##.

I get

##N \int_0^N \frac{1}{1+x^2} dx - \int_0^N \frac{x}{1+x^2} dx
\iff N\int_0^N \frac{1-x}{1+x^2} dx ##
How does one incorporate the ##N## in the integral and turn this into ##\arctan##?
 

fresh_42

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I do not understand your questions, sorry.
To obtain this expression in the first place, would it be reasonable to find an upper bound to the upper Riemann sum? Does it exist?
What does this mean? What expression? And why do you want to find an upper bound for something which is already an upper bound for something else? Why? What for?

In order ##N^\alpha \sum_{n=1}^N \operatorname{arctan} n## to converge, the ##\alpha## has to be so small, that it compensates the ##N\cdot\frac{\pi}{2}## term of the ##\operatorname{arctan}## as well as the increasing ##N##.
By the way, out of curiosity, how does one get ##\int_0^N \arctan{(x)} dx## from ##N \int_0^N \frac{1}{1+x^2} dx - \int_0^N \frac{x}{1+x^2} dx##.

I get

##N \int_0^N \frac{1}{1+x^2} dx - \int_0^N \frac{x}{1+x^2} dx
\iff N\int_0^N \frac{1-x}{1+x^2} dx ##
How does one incorporate the ##N## in the integral and turn this into ##\arctan##?
I was talking about $$\int \operatorname{arctan}\frac{x}{a}\,dx = x\operatorname{arctan}\frac{x}{a}-\frac{a}{2}\log(a^2+x^2)$$ and not about the derivative.

I'm not sure how to differentiate ##\operatorname{arctan}##. I would try the inverse function theorem or a Weierstraß substitution to integrate ##\frac{1}{1+x^2}##.
 
Since

##N \arctan{(N)}-(1/2)\ln{(1+N^2)}=\int_0^N \arctan{(x)} dx##​

but ##N \arctan{(N)}-(1/2)\ln{(1+N^2)}## is also equal to

##N \int_0^N \frac{1}{1+x^2} dx - \int_0^N \frac{x}{1+x^2} dx = N\int_0^N \frac{1-x}{1+x^2} dx##
How does one get that ##N\int_0^N \frac{1-x}{1+x^2} dx =\int_0^N \arctan{(x)} dx## ?
 

fresh_42

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Why is ##\pi \,\int_0^{\pi} \frac{4x}{\pi^3}\,dx = \int_0^{\pi} \operatorname{sin}x\,dx\,?##

Finite integrals are just a real number. So why are two different descriptions of a real number equal? Of course as ##\frac{d}{dx}\operatorname{arctan}x = \frac{1}{1+x^2} ## there are connections between those expressions. My suspicion is that this will lead to the Weierstraß substitution.
 
Okay. So ##N\int_0^N \frac{1-x}{1+x^2} dx## is not necessarily equal to ##\int_0^N \arctan{(x)} dx##?
In order ##N^\alpha \sum_{n=1}^N \operatorname{arctan} n## to converge, the ##\alpha## has to be so small, that it compensates the ##N\cdot\frac{\pi}{2}## term of the ##\operatorname{arctan}## as well as the increasing ##N##.
Can there possibly be any ##\alpha## that compensates for this? Won't ##N^{\alpha}## yield ##[\infty]^{\alpha}## as ##N\to\infty##? If ##\alpha<0##, the limit ##\beta## will be 0 (but ##\beta\neq0##), and for all other ##\alpha## the limit will be infinite, or?
 

fresh_42

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Okay. So ##N\int_0^N \frac{1-x}{1+x^2} dx## is not necessarily equal to ##\int_0^N \arctan{(x)} dx##?
It is equal, you can calculate both sides. But what does it mean if two formulas yield the same result? In our case there are connections between ##\operatorname{arctan}x ## and ##\frac{1}{x^2+1}## but not obvious ones. As mentioned, I think the Weierstaß substitution is a good place for further investigations.
Can there possibly be any ##\alpha## that compensates for this? Won't ##N^{\alpha}## yield ##[\infty]^{\alpha}## as ##N\to\infty##? If ##\alpha<0##, the limit ##\beta## will be 0 (but ##\beta\neq0##), and for all other ##\alpha## the limit will be infinite, or?
Write it down as your result. However, you have forgotten a case!
 

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