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- Find the exact sum of each series by matching it with a series in the table.\displaystyle\sum_{n=1}^{\infty} \dfrac{2^n}{n!}\displaystyle\sum_{n=0}^{\infty} \dfrac{(-1)^n}{(2n)!}\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n\,2^n}\displaystyle\sum_{n=0}^{\infty} \dfrac{(-1)^n}{2n+1}\displaystyle\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}\,2^n}{n}
- Let $f
- Let $f
- Let $f
- For each function, build the Maclaurin series from the table.
- Decide whether each function has a Maclaurin series.
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The common Maclaurin series: the lesson ›