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

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The discussion focuses on determining the intervals and radii of convergence for two series. For the series $\sum_{n=1}^{\infty} \frac{6x^n}{\sqrt[5]{n}}$, the radius of convergence is found to be 1, with convergence for $x \in (-1, 1)$, requiring further checks at the endpoints. The second series, $\sum_{n=1}^{\infty} \frac{8^n x^n}{(n+5)^2}$, has a radius of convergence of 1/8. There is confusion regarding the notation in the first series, which is clarified to be a typo, confirming it should be indexed by n. The discussion emphasizes the importance of correctly identifying series notation for accurate convergence analysis.
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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 $1$.

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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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