Which of the following series converge?

• \(\displaystyle \sum_{n=1}^\infty \frac{8^n}{n!} \)
• \(\displaystyle \sum_{n=1}^\infty \frac{n!}{n^{100}} \)
• \(\displaystyle \sum_{n=1}^\infty \frac{n+1}{(n)(n+2)(n+3)} \)

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We can use the ratio test for the first two options. To perform the ratio test, we find: $$ \lim_{n\to \infty} \left| \frac{a_{n+1}}{a_n} \right| $$

If the limit is less than 1, then the series converges absolutely. The series diverges if the value of the limit approaches infinity. It is inconclusive if it equals 1.

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$$ \sum_{n=1}^\infty \frac{8^n}{n!}$$ $$ \lim_{n\to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n\to \infty} \frac{8^{n+1}}{(n+1)!}\cdot \frac{n!}{8^n} $$ $$ =\lim_{n\to \infty} \frac{8}{n+1} = 0 \lt 1 \hskip{2em} \text{converges} $$ $$ \sum_{n=1}^\infty \frac{n!}{n^{100}} $$ $$ \lim_{n\to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n\to\infty} \frac{(n+1)!}{(n+1)^{100}} \cdot\frac{n^{100}}{n!} $$ $$ = \lim_{n\to\infty} \frac{n+1}{100} = \infty \hskip{2em} \text{diverges} $$

Use the limit comparison test for the last option:

$$ \sum_{n=1}^\infty \frac{n+1}{(n)(n+2)(n+3)} $$

We can compare the series with the convergent \(p\)-series \((p=2)\), \(b_n=\dfrac{1}{n^2}\)

$$ \lim_{n\to\infty} \frac{a_n}{b_n} = \frac{n+1}{(n)(n+2)(n+3)}\cdot \frac{n^2}{1} $$ $$ = \lim_{n\to\infty} = \frac{n^3+n^2}{n^3+6n^2+6} = 1 $$

Therefore, both series converge.