## What is the absolute convergence set and the radius of convergence of the next series?

MULTIPLE SELECTION ITEMS WITH THE SINGLE ANSWER You will then find questions that are developed around a statement, problem or context, against which you must select the option that correctly answers the question posed by four identified with the letters A, B , C, D. Once you select it, mark it with an oval that corresponds and justify the answer. [] (// blob.todoexpertos.com/uploads/md/c4cbd89df89ebc02a59ed56325deb4c0.jpg)

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You can make use of the criterion of reason, then, 
\begin{align}&\lim_{n\rightarrow\infty}{\left|\frac{a_{n+1}}{a_{n}}\right|}=\lim_{n\rightarrow\infty}{\left|\frac{\frac{x^{n+1}}{\sqrt{n+1}}}{\frac{x^{n}}{\sqrt{n}}}\right|}=\lim_{n\rightarrow\infty}{\left|\frac{x^{n+1}}{\sqrt{n+1}}\frac{\sqrt{n}}{x^{n}}\right|}=\lim_{n\rightarrow\infty}{|x|\left|\frac{\sqrt{n}}{\sqrt{n+1}}\right|}=|x|\lim_{n\rightarrow\infty}{\left|\sqrt{\frac{n}{n+1}}\right|}\\&\\&...=|x|\lim_{n\rightarrow\infty}\sqrt{1-\frac{1}{n+1}}=|x|(1)=|x|<1\\&\\&\end{align}if we do all of that is strictly less than one, we guarantee that the series converges absolutely...so,


\begin{align}&|x|<1\hspace{5mm}\textrm{is equivalent to: }\hspace{5mm} -1< x<1\end{align}is set then the frayed of convergence is 1, and if we want to know the interval of convergence, we have to analyze, which happens with the series right at the points of intersection of the ball is say -1 and 1, then


\begin{align}&: x=1\\&\sum_{n=0}^{\infty}{\frac{1^{n}}{\sqrt{n}}}\\&\textrm{if we analyze this limit, you will notice that it is divergent, now let }\\&\\&For:x=-1\\&\sum_{n=0}^{\infty}{\frac{(-1)^{n}}{\sqrt{n}}}\\&\textrm{if we analyze this limit, you will notice that it is convergent}\\&\\&\end{align}verify these limits, your remaining task...you have to find the limit when n tends to infinity...and well, that would be all...then the interval of convergence, it would be,


\begin{align}&R:[-1,1)\end{align}

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