The convergence of a numerical series

[Amaelle]

Homework Statement

Look at the image

Relevant Equations

Absolute convergence.

Greetings
here is the exercice

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

I do unsderstand it very well but I need to know where my logics has failed me in my attempt
thank you!

 

[FactChecker]

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.

 

[pasmith]

You can make the non-convergence rigorous by using the inequality $$\sin x > \frac{2}{\pi} x, \qquad 0 < x < \frac\pi 2.$$ Then $$\sum_{n=0}^N \sin\left(\frac{\pi}{n+1}\right) > \sin\left(\frac{\pi}1\right) + \sum_{n=1}^N \frac{2}{n+1} = 2\sum_{n=2}^{N+1} \frac 1n$$ which diverges.

 

[Mark44]

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).

 

[Amaelle]

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

 

[FactChecker]

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 $+\infty$.