# Chapter 10: Infinite Sequences and Series - Section 10.4 - Comparison Tests - Exercises 10.4 - Page 591: 43

Converges

#### Work Step by Step

Let $u_n=\Sigma_{n=2}^\infty \dfrac{1}{n !}$ Re-write the given series such as: $\Sigma_{n=2}^\infty \dfrac{1}{n !}=\Sigma_{n=2}^\infty \dfrac{1}{n (n-1)(n-2)(n-3)!}$ We compare with: $\Sigma_{n=2}^\infty \dfrac{1}{n (n-1)!} \lt \Sigma_{n=2}^\infty \dfrac{1}{(n-1)(n-1)}=\Sigma_{n=1}^\infty (\dfrac{1}{n^2})$ This implies that $\Sigma_{n=2}^\infty (\dfrac{1}{n !}) \lt \Sigma_{n=1}^\infty (\dfrac{1}{n^2})$ and $\Sigma_{n=1}^\infty \dfrac{1}{n^2}$ shows a convergent $p$-series with $p=2 \gt 1$ . Thus, the series converges by the limit comparison test.

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