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For the infinite series, $$ \sum_{n=1}^{\infty}\frac{x^n(n+1)^{3n}}{(3n+1)!} $$

It is convergent if $$x\in \left [ -\frac{3^3}{e^3},\frac{3^3}{e^3} \right ]$$

I used the Ratio Rule in this question, but my answer came out to be that $x$ should be in between $+27$ and $-27$, missing the $e^3$. Can someone tell me how this question should be dealt with?

BX3129
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    You should show us how you conducted the ratio test – FShrike Mar 01 '23 at 10:33
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    Are you aware of the fact that $(1 + \frac{1}{n})^n \to e$ as $n \to \infty$? It's a common mistake to think that it tends to $1$; see here, for example: https://math.stackexchange.com/questions/136784/why-lim-limits-n-to-infty-left1-frac1n-rightn-doesnt-evaluate-to – Hans Lundmark Mar 01 '23 at 10:34
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    Thanks for the hints guys , my problem was that I calculated $\left(\frac{n+2}{n+1}\right)^{3n}$ into $1$ instead of turning it into $\left(1+\frac{1}{n+1}\right)^{3n}$, which clearly wasn't right. – BX3129 Mar 01 '23 at 11:27

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Be careful how you do the ratio test. Letting $$a_n=\frac{\vert x \vert^n(n+1)^{3n}}{(3n+1)!},$$ we have $$\frac{a_{n+1}}{a_n}=\vert x \vert(\frac{n+2}{n+1})^{3n}\frac{(n+2)^3}{(3n+4)(3n+3)(3n+2)}$$ When you take the limit of $\frac{a_{n+1}}{a_n}$ as $n \to \infty$, the only hard part is the limit of $$(\frac{n+2}{n+1})^{3n}$$ as $ n \to \infty$. Take natural logarithms and work out the limit of $$\frac{\ln(n+2)-\ln(n+1)}{\frac{1}{3n}}$$ by l'Hospital's Rule. Then raise this limit to the power $e$ and multiply by the other limit.

P. Lawrence
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