A problem with the convergence of a series

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The discussion revolves around the convergence of a series and the application of the root test, which indicates convergence with a limit of 2/3. However, confusion arises regarding the requirement for the sequence's terms to approach zero, as the limit of ((2n+100)/(3n+1))^n approaches infinity. Participants clarify that the logarithm's denominator, 3n+1, must be considered in the calculations. After re-evaluating the approach, the original poster recognizes their mistake. The conversation concludes with a resolution of the confusion regarding the convergence criteria.
Amaelle
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Homework Statement
Show that the following sequence is convergent
Relevant Equations
Racine test
Good day
I have a question about the convergence of the following serie
inf.png


I understand that the racine test shows that it an goes to 2/3 which makes it convergent
but I also know that for a sequence to be convergent the term an should goes to 0 but the lim(n---->inf) ((2n+100)/(3n+1))^n)=lim exp(n*log(2n+100)/(3n+1))=+infinity
I'm really confused
thank you!
 
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What do you get if you cancel the quotient by ##n## instead of taking the logarithm?
 
Amaelle said:
lim exp(n*log(2n+100)/(3n+1))=+infinity
You should rethink this part. Try plugging in a large value of ##n## and see if it's what you expect.
 
Amaelle said:
Homework Statement:: Show that the following sequence is convergent
Relevant Equations:: Racine test

Good day
I have a question about the convergence of the following serie
View attachment 283529

I understand that the racine test shows that it an goes to 2/3 which makes it convergent
but I also know that for a sequence to be convergent the term an should goes to 0 but the lim(n---->inf) ((2n+100)/(3n+1))^n)=lim exp(n*log(2n+100)/(3n+1))=+infinity
I'm really confused
thank you!
The denominator, 3n+1, is also inside the log.
 
vela said:
You should rethink this part. Try plugging in a large value of ##n## and see if it's what you expect.
thank you very much I just spotted the mistake!
 
FactChecker said:
The denominator, 3n+1, is also inside the log.
thank you it's clear now
 
First, I tried to show that ##f_n## converges uniformly on ##[0,2\pi]##, which is true since ##f_n \rightarrow 0## for ##n \rightarrow \infty## and ##\sigma_n=\mathrm{sup}\left| \frac{\sin\left(\frac{n^2}{n+\frac 15}x\right)}{n^{x^2-3x+3}} \right| \leq \frac{1}{|n^{x^2-3x+3}|} \leq \frac{1}{n^{\frac 34}}\rightarrow 0##. I can't use neither Leibnitz's test nor Abel's test. For Dirichlet's test I would need to show, that ##\sin\left(\frac{n^2}{n+\frac 15}x \right)## has partialy bounded sums...

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