The convergence of a numerical series

Amaelle
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Homework Statement
Look at the image
Relevant Equations
Absolute convergence.
Greetings
here is the exercice
1629802637709.png

My solution was
as n^2+n+1/(n+1) tends asymptotically to n then the entire stuffs inside the sinus function tends to npi which make it asymptotically equal to sin(npi) which is equal to 0 and consequently the sequence is Absolutely convergent

Here is the solution of the book
1629802909490.png

1629802963524.png


I do unsderstand it very well but I need to know where my logics has failed me in my attempt
thank you!
 
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To say that the terms are asymptotically equal to something does not say enough. It doesn't say how fast it approaches that limit. For instance, ##\sum{1/n}## doesn't converge even though ##1/n## approaches 0 asymptotically.
 
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You can make the non-convergence rigorous by using the inequality <br /> \sin x &gt; \frac{2}{\pi} x, \qquad 0 &lt; x &lt; \frac\pi 2. Then \sum_{n=0}^N \sin\left(\frac{\pi}{n+1}\right) &gt; \sin\left(\frac{\pi}1\right) + \sum_{n=1}^N \frac{2}{n+1} = 2\sum_{n=2}^{N+1} \frac 1n which diverges.
 
Amaelle said:
Homework Statement:: Look at the image
Relevant Equations:: Absolute convergence.

My solution was
as n^2+n+1/(n+1)
You need more parentheses here. Taken literally, the above means
##n^2 + n + \frac 1{n + 1}##, which you surely didn't intend.
If you don't use Tex, it should be written as (n^2 + n + 1)/(n + 1).
 
FactChecker said:
To say that the terms are asymptotically equal to something does not say enough. It doesn't say how fast it approaches that limit. For instance, ##\sum{1/n}## doesn't converge even though ##1/n## approaches 0 asymptotically.

I wanted to say as
sin[pi*((n^2+n+1)/(n+1)]≈sin[pi*((n^2)/(n)]≈sin[pi(n)] and we know the serie ∑sin[pi(n)] converges SO IS MY SERIE
 
Once again, your "##\approx##" is not enough. It depends on how fast the terms on the left come close to the terms on the right. Consider ##1/n \approx 0## as ##n## gets large. But ##\sum{0}## converges while ##\sum{1/n}## diverges to ##+\infty##.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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