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Determine whether the series converges or diverges.

$$ \sum _{n=1}^{\infty }\:\left(\frac{19}{n!}\right) $$

I know that this question a lot easier if I use ratio test but I have not learned ratio test yet. The only option I have is divergence, comparison, limit comparison, and integral test. How can I prove that this series converges by using the limited tests.

Thanks in advance.

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    $\begingroup$ Hint: $\dfrac{19}{n!}\le \dfrac{n}{n!}=\dfrac{1}{(n-1)!}$ for all $n\ge 19$. $\endgroup$ – user 170039 Jul 2 '17 at 14:01
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$$\sum _{n=1}^{\infty }\frac{19}{n!}=19\sum _{n=1}^{\infty }\frac{1}{n!}=19\Bigl(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dotsb\Bigr)\\= 19\Bigr(-\frac{1}{0!}+\underbrace{\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dotsb}_{e}\Bigr)=19(-1+e)$$

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Then use the fact that $(\forall n\in\mathbb{N}\setminus\{2,3\}):\frac{19}{n!}\leqslant\frac{19}{n^2}$ and apply the integral test in order to prove that $\sum_{n=1}^\infty\frac{19}{n^2}$ converges.

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  • $\begingroup$ The inequality fails for $n=2,3.$ $\endgroup$ – zhw. Jul 2 '17 at 15:27
  • $\begingroup$ @zhw. Corrected. Thanks. $\endgroup$ – José Carlos Santos Jul 2 '17 at 15:29
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$$\sum_{n=1}^{\infty}\frac{19}{n!}=19\left(1+\sum_{n=2}^{\infty}\frac{1}{n!}\right)<19\left(1+\sum_{n=1}^{\infty}\frac{1}{n(n+1)}\right)=$$ $$=19\left(1+\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+1}\right)\right)=19(1+1)=38.$$ Thus, our series converges.

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Forget about the constant $19.$ Note that in $n!$ you have $n-1$ factors, each of which is $\ge 2.$ Thus $n! \ge 2^{n-1} \implies 1/n! \le 1/2^{n-1}.$ Use the comparison test.

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Use inequality $$ n! > 2^n, \ n\geqslant 4.$$ Then for $n\geqslant 4$ $$\frac{1}{n!} < \frac{1}{2^n},$$ so $$\sum_{n=1}^{\infty }\frac{19}{n!} = 19\sum_{n=1}^{\infty }\frac{1}{n!} < 19\left(1 + \frac{1}{2} + \frac{1}{6} + \sum_{n=4}^{\infty }\frac{1}{2^n} \right)=19\left(1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{8}\right).$$

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