Finding the Radius of Convergence for Y=6x+16 - Troubleshooting and Solution

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The discussion revolves around finding the radius of convergence for the function Y=6x+16. A participant initially calculated the radius to be 5/3 but later questioned the accuracy of this result. There is confusion regarding the notation used for logarithmic expressions and how it relates to convergence criteria. Participants also discuss the importance of defining the variable t in relation to series convergence. The thread highlights the need for clarity in notation and understanding the underlying rules for determining convergence.
Amaelle
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Homework Statement
Loook at the image
Relevant Equations
Power series.
Raduis of convergence.
Greetings
I have some problems finding the correct result
1629716024934.png

My solution:
I puted Y=6x+16
so now will try to find the raduis of convergence of Y
so let's calculate the raduis criteria of convergence:
1629716182475.png

1629716392658.png

  • We know that Y=6x+16
  • Conseqyently -21/6<=x<=-11/6 so the raduis must be 5/3. But this is not the solution!
Here is the solution of the book
1629716620401.png

1629716656750.png


I would like to know where is my mistake
thank you!
 
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I am not sure about your notation \log^{2021}x. Is it
\log_{2021}x or
(\log x)^{2021}?
 
5 for 6x+16 and 5/2 for 3x+8 seem not different. What is the rule to define t for t^n series for convergence ?
 
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anuttarasammyak said:
I am not sure about your notation \log^{2021}x. Is it
\log_{2021}x or
(\log x)^{2021}?
(\log x)^{2021}?
 
anuttarasammyak said:
5 for 6x+16 and 5/2 for 3x+8 seem not different. What is the rule to define t for t^n series for convergence ?

I realize that I have the same confusion, I thought we were looking for the domain where x is convergent?
I understand I got the same results implicitely but did the book go for that particular set up?
thanks a million!
 

Thread 'Does this series converge uniformly?'

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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