Evaluate the infinite sum of 2^n / n! for n from 0 to infinity.

Step-by-step solution

1. 1

   Confirm the series converges before summing it. By the ratio test,

   $\frac{a_{n+1}}{a_{n}}=\frac{2^{n+1}/(n+1)!}{2^{n}/n!}=\frac{2}{n+1}\longrightarrow 0<1$

   so the series converges absolutely for this and indeed for any base in place of $2$ — factorials always beat geometric growth eventually.

2. 2

   Match the series to a known expansion. The exponential function has the Maclaurin series

   $e^{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!},\qquad \text{valid for all real }x$

   The given series is exactly this with $x=2$, so no summation work is needed — only recognition.

3. 3

   Read off the value.

   $\sum_{n=0}^{\infty}\frac{2^{n}}{n!}=e^{2}\approx 7.389056$

4. 4

   Watch the starting index. If the sum begins at $n=1$ instead, the $n=0$ term $\tfrac{2^0}{0!}=1$ is missing:

   $\sum_{n=1}^{\infty}\frac{2^{n}}{n!}=e^{2}-1\approx 6.389056$

   This is the one detail that decides the answer, and it is worth stating explicitly whenever the lower limit is not written clearly.

5. 5

   Check by partial sums. Adding the first terms: $1+2+2+1.3333+0.6667+0.2667+0.0889+0.0254=7.3809$, already within $0.008$ of $e^2=7.389056$. The remaining tail is dominated by $\tfrac{2^8}{8!}=0.00635$, matching the observed gap $\checkmark$.