Discussion

The sum of the following infinite series is:

\[ \begin{aligned} &\frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} \\ &+ \frac{1}{4!} + \frac{1}{5!} + \cdots \end{aligned} \]

Answer: C

The sum is \(e - 1\).

The exponential series is

\[ \begin{aligned} e &= \frac{1}{0!} + \frac{1}{1!} \\ &+ \frac{1}{2!} + \frac{1}{3!} + \cdots \end{aligned} \]

The given series is the same except that it starts at \(\frac{1}{1!}\), so it leaves out the first term \(\frac{1}{0!} = 1\):

\[ \sum_{n=1}^{\infty} \frac{1}{n!} = e - 1 \approx 1.718 \]

\(e\) would include the missing 1, \(1 + e\) adds it instead of removing it, and \(\pi\) has nothing to do with this series.

No comment is present. Be the first to comment.

Loading…