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Does the series $\displaystyle\sum_{n=1}^{\infty} \dfrac{19+(n+5)!}{(n+7)!}$ Converge or diverge

I tried using the ratio test, but it gave me this which I don't see how you'd be able to solve. $$\lim_{n\to\infty} |\frac{a_{n+1}}{a_{n}}| = \lim_{n\to\infty}\frac{19+(n+6)!}{19+(n+5)!*(n+8)}$$

Wolfram alpha told me to use the comparison test, but I can't for the love of god see which series I'd compare it to.

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    $\begingroup$ Divide the numerator and denominator by $(n+5)!$. Ultimately you’ll have a series like $1/n^2$. The ratio test won’t work to show convergence, but the limit comparison test will. $\endgroup$ – Joel Nov 8 '17 at 14:41
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    $\begingroup$ Yes like @Joel writes you can probably squeeze it with some $\sum\frac{1}{an^2+bn+c}$ $\endgroup$ – mathreadler Nov 8 '17 at 14:43
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    $\begingroup$ Yeah, that works guys. Thank you for the help. It was easier than it seemed. I must've gone blind while thinking about it for too long. $\endgroup$ – Andreas Nov 8 '17 at 14:48
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$$\sum_{n=1}^{\infty} \frac{19+(n+5)!}{(n+7)!}=19\sum_{n=1}^{\infty} \frac{1}{(n+7)!}+\sum_{n=1}^{\infty} \frac{1}{(n+7)(n+6)}$$ and $$ \frac{1}{(n+7)!}<\dfrac{1}{n^2}~~~,~~~\frac{1}{(n+7)(n+6)}<\dfrac{1}{n^2}$$

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$$\sum_{n=1}^\infty\frac{19+(n+5)!}{(n+7)!}=\sum_{n=1}^\infty\frac{1+\frac{19}{(n+5)!}}{(n+6)(n+7)}<\sum_{n=1}^\infty\frac2{n^2}$$

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