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I need to find all $x$ such that the series $\sum_{n=0}^\infty \frac{x^nn^2}{(2n)\cdots (6)(4)(2)}$ converges using the ratio test.

We apply the ratio test and evaluate the limit

$$ \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} = \frac{x^{n+1}(n+1)^2}{(2n+2)(2n)\cdots(6)(4)(2)} \frac{(2n)\cdots(6)(4)(2)}{x^nn^2}= \frac{x(n+1)}{2n^2}$$

I need this limit to be less than 1 for my series to converge, so the $x$ value needs to bring this fraction below $1$? I need some pointers. Thank you!

Edit: Okay I have an idea... actually evaluating the limit. Using L'Hopital's rule we can say that the limit of the final expression is 0, meaning that this will converge for all $x$?

MCMath
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    Since you wrote $ \lim_{n \rightarrow \infty}$ on the left, you must write it also on the middle and on the right. I prefer to write it nowhere. – Anne Bauval Nov 18 '24 at 23:20

2 Answers2

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Hint: you don't need anything complicated to evaluate the limit, just carry on with some more algebraic simplification to separate concerns and reduce to limits you should know:

$$\frac{x(n+1)}{2n^2} = \frac{1}{2}\left(\frac{x}{n} + \frac{x}{n^2}\right)$$

Can you take it from there?

$\frac{x}{n}$ and $\frac{x}{n^2}$ both tend to zero as $n$ tends to infinity for any $x$. So the series converges for any $x$.

Rob Arthan
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  • I can certainly try. We need it to be $<1$, so $\frac{x}{n} + \frac{x}{n^2} < 2$. But the limit is taking $n$ to infinity and therefore those will be 0... so the $x$ value doesn't matter? – MCMath Nov 19 '24 at 02:06
  • I'm not sure my ratio is correct though... the sequence on the bottom might not be correct? – MCMath Nov 19 '24 at 02:16
  • @MCMath: the value of $x$ doesn't matter. I've added some more info under a spoiler cover that should help you get the solution. – Rob Arthan Nov 19 '24 at 03:16
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Indeed "the limit of the final expression is $0$, meaning that this will converge for all $x$", but L'Hopital is useless here: as $n\to\infty$, $$\frac{a_{n+1}}{a_n}=\frac x2\frac{n+1}{n^2}=\frac x2\frac{1+\frac1n}n\to0.$$ For a generalization of this method, see How to know if it diverges or converges and finding the convergent value.

Anne Bauval
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  • Is it necessary to say that $x$ should be finite or does the multi variable limit always evaluate to $0$ when $n \to \infty$? Sorry I am not familiar with assumptions when proving convergence. – Srini Nov 18 '24 at 23:28
  • $x$ was assumed to be finite (real, or possibly complex). – Anne Bauval Nov 18 '24 at 23:29
  • @AnneBauval L'Hopital's rule cannot be used for infinite limits? – MCMath Nov 19 '24 at 02:07
  • l'Hopital's rule does work for $\frac{n+1}{n^2}$ and reduces it to $\frac{1}{n}$ which tends to $0$ as $n$ tends to $\infty$. However, I think it's simpler to solve this without l'Hopital's rule. – Rob Arthan Nov 19 '24 at 03:23
  • @MCMath I said it was useless, not that it cannot be used. In this example, it is undoubtedly overkill. Better learn how to get rid of such simple indeterminate forms $\frac\infty\infty$ by the usual method shown in my answer: factorization and cancellation. – Anne Bauval Nov 19 '24 at 07:15
  • Why the downvote? – Anne Bauval Nov 19 '24 at 07:17