`sum_(n=0)^oo 1/(n!)` Use the Direct Comparison Test to determine the convergence or divergence of the series.

`n! gt n^2 ` for all n > 3, so except for the first few terms, this series is strictly bounded above by the p-series 1/n^2:

`sum_{4}^{infty} 1/{n!} lt sum_{4}^{infty} 1/n^2`

Finitely many terms won't affect convergence, so since the p-series converges, the `1/n!` series must also converge by direct comparison.

Comments

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