Problem studying the convergence of a series

[Amaelle]

Homework Statement

Convergence of a serie (look at the image)

Relevant Equations

absolute convergence, taylor expnasion, asymptotic behaviour!

Good day
here is the exercice

and here is the solution that I understand very well

but I have a confusion I hope someone can explain me
if I take the taylor expansion of sin ((n^2+n+1/(n+1))*pi)≈n^2+n+1/(n+1))*pi≈n*pi which diverge!
I know something is wrong in my logic please help me

many thanks in advance!
best regards!

 

[fresh_42]

Homework Statement:: Convergence of a serie (look at the image)
Relevant Equations:: absolute convergence, taylor expnasion, asymptotic behaviour!

if I take the taylor expansion of sin ((n^2+n+1/(n+1))*pi)≈n^2+n+1/(n+1))*pi≈n*pi which diverge!

This is what you get by a quick, but wrong look at it. To cut off all the rest of the Taylor series gives you a false impression. With this method you either have to use the complete series or investigate merely absolute convergence, in which case you need the triangle inequality.

Anyway, the point is, that a) $\sin(x)\approx x$ only holds for $x\ll 1$ and b) that we actually have $\sin(f(n)\cdot \pi )$ with $f(n)=\dfrac{n^2+n+1}{n+1}$ which itself converges to $n\pi$ and $\sin(n\pi)=0$.

Hence we add numbers which are close to zero. We know from $\sum \dfrac{1}{n} \to \infty$, $\sum \dfrac{1}{n^2} \to \pi^2/6$, and $\sum (-1)^n\dfrac{1}{n} \to -\log(2)$ that in such cases we have to have a close look on "how close to zero". This means that we must keep equality as long as possible before we can start to simplify by Taylor or other means. That is where the addition theorem comes into play.

It is generally a good advice for any calculation: Stay accurate as long as possible and turn to estimations and approximations at the very last.

 

[anuttarasammyak]

Homework Statement:: Convergence of a serie (look at the image)
Relevant Equations:: absolute convergence, taylor expnasion, asymptotic behaviour!

if I take the taylor expansion of sin ((n^2+n+1/(n+1))*pi)≈n^2+n+1/(n+1))*pi≈n*pi which diverge!

You seem to take only first order of Taylor expansion. It is no good because
$$\frac{n^2+n+1}{n+1}=n+\frac{1}{n+1}>1$$
We should remove $n$ to apply Taylor series properly as the text does.

 

[Amaelle]

This is what you get by a quick, but wrong look at it. To cut off all the rest of the Taylor series gives you a false impression. With this method you either have to use the complete series or investigate merely absolute convergence, in which case you need the triangle inequality.

Anyway, the point is, that a) $\sin(x)\approx x$ only holds for $x\ll 1$ and b) that we actually have $\sin(f(n)\cdot \pi )$ with $f(n)=\dfrac{n^2+n+1}{n+1}$ which itself converges to $n\pi$ and $\sin(n\pi)=0$.

Hence we add numbers which are close to zero. We know from $\sum \dfrac{1}{n} \to \infty$, $\sum \dfrac{1}{n^2} \to \pi^2/6$, and $\sum (-1)^n\dfrac{1}{n} \to -\log(2)$ that in such cases we have to have a close look on "how close to zero". This means that we must keep equality as long as possible before we can start to simplify by Taylor or other means. That is where the addition theorem comes into play.

It is generally a good advice for any calculation: Stay accurate as long as possible and turn to estimations and approximations at the very last.

thanks a million it's clear now!

 

[Amaelle]

You seem to take only first order of Taylor expansion. It is no good because
$$\frac{n^2+n+1}{n+1}=n+\frac{1}{n+1}>1$$
We should remove $n$ to apply Taylor series properly as the text does.

thanks a million for this nice shot!