What are the intervals and radius of convergence for two series with typos?

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Discussion Overview

The discussion revolves around determining the intervals and radii of convergence for two series, with a focus on the application of convergence tests. The conversation includes both theoretical and practical aspects of series convergence.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant initially presents two series for analysis, including $\sum_{x=1}^{\infty} \frac{6x^n}{\sqrt[5]{n}}$ and $\sum_{n=1}^{\infty} \frac{8^nx^n}{(n+5)^2}$.
  • Another participant suggests using the integral test for the first series and the ratio test for the second but expresses uncertainty about the application of these tests.
  • A later post provides calculations for the radius of convergence for the second series, concluding it is $R=\frac{1}{8}$, but does not clarify the interval of convergence.
  • There is confusion regarding the first series, with one participant correcting their earlier interpretation, indicating it should be $\sum_{n=1}^{\infty} \frac{6x^n}{\sqrt[5]{n}}$ instead of $\sum_{x=1}^{\infty} \frac{6x^n}{\sqrt[5]{n}}$.
  • Another participant questions how to apply the ratio test to the corrected series.
  • One participant acknowledges the typo and states they figured out the series correctly.

Areas of Agreement / Disagreement

Participants express uncertainty and confusion regarding the correct interpretation of the series and the application of convergence tests. There is no consensus on the interval of convergence for the first series, and the discussion remains unresolved regarding the application of the ratio test.

Contextual Notes

The discussion highlights potential typos in the series definitions, which may affect the application of convergence tests. The participants do not fully resolve the implications of these typos on the convergence analysis.

ineedhelpnow
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find the interval of convergence of the series $\sum_{x=1}^{\infty} \frac{6x^n}{\sqrt[5]{n}}$find the radius of convergence of the series $\sum_{n=1}^{\infty} \frac{8^nx^n}{(n+5)^2}$
 
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for the first one i think you use the integral test and for the second the ratio test but i don't understand HOW you use them to determine the interval and radius?
 
ineedhelpnow said:
find the interval of convergence of the series $\sum_{x=1}^{\infty} \frac{6x^n}{\sqrt[5]{n}}$

$$\xi=0$$

$$\rho=\lim_{n \to +\infty} \sqrt[n]{|a_n|}=\lim_{n \to +\infty} \sqrt[n]{\frac{6}{\sqrt[5]{n}}}=\lim_{n \to +\infty} \frac{6^{\frac{1}{n}}}{n^{\frac{1}{5n}}}=1$$

$$R=\frac{1}{ \rho}=1$$

So,the series converges absolutely for $x \in (-1,1)$ and diverges for $x \notin [-1,1]$

Now you have to check the convergence at the points $-1$and $1$.

- - - Updated - - -

ineedhelpnow said:
find the radius of convergence of the series $\sum_{n=1}^{\infty} \frac{8^nx^n}{(n+5)^2}$

$$\xi=0$$

$$\rho=\lim_{n \to +\infty} \sqrt[n]{|a_n|}=\lim_{n \to +\infty} \sqrt[n]{\frac{8^n}{(n+5)^2}}=\lim_{n \to +\infty} \frac{8}{(n+5)^{\frac{2}{n}}}=8$$So,the radius of convergence is $R=\frac{1}{ \rho}=\frac{1}{8}$ .
 
thanks! what about the x though?
 
ineedhelpnow said:
thanks! what about the x though?

Oh,sorry! (Blush) I thought that it would be $\sum_{n=1}^{\infty} \frac{6x^n}{\sqrt[5]{n}}$.
When it is like that: $\sum_{x=1}^{\infty} \frac{6x^n}{\sqrt[5]{n}}$ , the ratio test cannot be used!
 
(Tmi) do you know how to do it with x? :o
 
ineedhelpnow said:
(Tmi) do you know how to do it with x? :o

Hi!

What do you get if you apply the ratio test?

In other words, what is:
$$L=\lim_{n\to \infty} \left| \frac{\dfrac{6x^{n+1}}{\sqrt[5]{n+1}}}{\dfrac{6x^n}{\sqrt[5]{n}}} \right|$$
 
ineedhelpnow said:
find the interval of convergence of the series $\sum_{x=1}^{\infty} \frac{6x^n}{\sqrt[5]{n}}$

Is it maybe a typo and the series is $\sum_{n=1}^{\infty} \frac{6x^n}{\sqrt[5]{n}}$ ? (Thinking)
 
that actually is a typo but i figured them out. thanks though
 

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