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The original series is $$\sum^\infty_{n=1} (\sqrt{n+1}-\sqrt{n})2^nx^{2n}.$$ I used The absolute value of the ratio of the latter item to the former item to do this problem. $2\left|\frac{(\sqrt{n+2}+\sqrt{n+1})y}{\sqrt{n+1}+\sqrt{n}}\right|$ （$y=x^2$）I used it to replace $x^2$.

How can I continue to reduce the series to a formula with a constant and $|x^2|$ to get the radius of convergence?

Harish Chandra Rajpoot
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Dianqing
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    Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – Another User May 15 '23 at 12:30
  • @Kenny Wong Thanks for your edition – Dianqing May 15 '23 at 12:38
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    @Dianqing No problem. However, we're waiting for your own edit - to provide context about your own thoughts. – Kenny Wong May 15 '23 at 12:42
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    Have you tried the ratio test? If so, show us where you start having trouble. – MasB May 15 '23 at 13:04
  • @MasB I can't continue to reduce the equation when it comes to $\frac{(\sqrt{n+2}-\sqrt{n+1})2x}{\sqrt{n+1}-\sqrt{n}}$ – Dianqing May 15 '23 at 13:11
  • It is not $2x$ but $x^2$ in your last comment. Use a binomial expansion and long division (or Taylor series). It is simple. – Claude Leibovici May 15 '23 at 13:16
  • @Claude Leibovici Could you show the detailed step for me i just can't find why $x^2$ is right.And the method you told is out of my learning range.Still thanks to u anyway. – Dianqing May 15 '23 at 13:28
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    Everybody here is ready to help you but no one will do your homework. Show some working, please. – Claude Leibovici May 15 '23 at 13:37
  • Based on $\lim_{n\to +∞}\vert{\frac{a_{n+1}x^{n+1}}{a_n x^n}}=l\vert{x}$ I transformed it to $\frac{(\sqrt{n+2}-\sqrt{n+1})2^{n+1}x^{2n+1}}{(\sqrt{n+1}-\sqrt{n})2^{n}x^{2n}}$. But I can only get $2x$ not the $x^{2}$ – Dianqing May 15 '23 at 14:02
  • Are you online now?@ClaudeLeibovici – Dianqing May 15 '23 at 14:21
  • $x^{2n}$ becomes $x^{2(n+1)}=x^{2n+2}$ If you showed what you have done, we could have caught this error at the start of the conversation. – MasB May 15 '23 at 16:29
  • @MasB Thanks for your answer.But when i got $|\frac{(\sqrt{n+2}-\sqrt{n+1})2x^{2}}{(\sqrt{n+1}-\sqrt{n})}|$.I can't get a formula which is like to $l|x|$ – Dianqing May 16 '23 at 01:17
  • Let $y=x^2$, Apply the tools you know to find the radius of convergence for the resultant series in $y$. Then figure out what that radius tells you about the radius of convergence in the original series in $x$. – Paul Sinclair May 16 '23 at 14:39
  • @Dianqing Note that $$ \left| {2\frac{{\sqrt {n + 2} + \sqrt {n + 1} }}{{\sqrt {n + 1} + \sqrt n }}y} \right| \to 2|y| = 2|x|^2 $$ as $n\to+\infty$. Now you want this to be less than $1$ to ensure convergence (via ratio test). What does that give you for $|x|$? – Gary May 17 '23 at 06:58
  • @Gary Because my textbook told me I have to get $l|x|$(l is a constant) and use the public formula $R=\frac{1}{l}$ to get the radius of convergence. So is that $\frac{1}{\sqrt{2}}$ the radius,right? – Dianqing May 17 '23 at 13:06
  • @Dianqing Yes, it is. – Gary May 17 '23 at 13:19

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