Evaluate $\sum_{n=0}^{\infty} \frac{5^n}{n!}$

Key takeaway: This is one of those series that collapses into a familiar function immediately. If you know the Maclaurin series for $e^x$, the result is almost automatic.

We recognize the standard exponential series:

$e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!}.$

Here, substitute $x=5$:

$e^5=\sum_{n=0}^{\infty} \frac{5^n}{n!}.$

So the value of the series is

$\boxed{e^5}.$

Quick check

The factorial in the denominator guarantees convergence, so the infinite sum is well-defined and exactly matches the exponential function at $x=5$.

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Pitfalls the pros know 👇 A common mistake is treating this like a geometric series. It is not geometric because the denominator is $n!$, not a fixed power of a constant. Another slip is forgetting that the series starts at $n=0$, which matters for matching the exponential expansion exactly.

What if the problem changes? If the top were $(-5)^n$ instead, the sum would be $e^{-5}$. More generally, replacing 5 by any number $a$ gives $\sum_{n=0}^{\infty} a^n/n! = e^a$.

Tags: exponential series, Maclaurin series, factorial series