Prove that: (sum from n=0 to infinity frac1n!) (sum
Prove that: $$ \left(\sum_{n=0}^{\infty} \frac1{n!}\right)\left(\sum_{n=0}^{\infty} \frac{(-1)^n}{n!}\right)=1. $$
Explanation of Method
Recognizing both series as $e^x$ evaluated at $x=1$ and $x=-1$ gives the identity immediately via $e\cdot e^{-1}=1$. A second, more instructive proof forms the Cauchy product directly: the binomial theorem collapses the $n$-th product term into $(-1)^n0^n/n!$, which vanishes for every $n\ge1$ and equals $1$ only at $n=0$ — reproducing the same result from first principles without invoking the exponential function's known value.
- Identify both series as $e$ and $e^{-1}$ directly.
- Form the Cauchy product and simplify via the binomial theorem.
- Sum the resulting (mostly zero) product series to confirm the same value.
Step 1 · Direct identification
$\sum_{n=0}^{\infty}1/n!=e^1=e$ and $\sum_{n=0}^{\infty}(-1)^n/n!=e^{-1}$, so the product is $e\cdot e^{-1}=1$ immediately.
Step 2 · Cauchy product from first principles
With $a_n=1/n!$, $b_n=(-1)^n/n!$: $$ c_n=\sum_{k=0}^n\frac1{k!}\cdot\frac{(-1)^{n-k}}{(n-k)!} = \frac{(-1)^n}{n!}\sum_{k=0}^n\binom nk(-1)^k, $$ using $(-1)^{n-k}=(-1)^n(-1)^{-k}=(-1)^n(-1)^k$ and $\binom nk=n!/[k!(n-k)!]$.
Step 3 · Apply the binomial theorem
$\sum_{k=0}^n\binom nk(-1)^k=(1-1)^n=0^n$, which is $1$ for $n=0$ and $0$ for $n\ge1$. So $c_0=1$, $c_n=0$ for $n\ge1$, and since both factor series converge absolutely, the Cauchy product converges to the product of the sums: $\sum_{n=0}^{\infty}c_n=1$.
Summary Table
| Series | Value |
|---|---|
| $\sum1/n!$ | $e$ |
| $\sum(-1)^n/n!$ | $e^{-1}$ |
| Cauchy product term $c_n$ | $1$ if $n=0$, else $0$ |
Additional Notes
- This is the cleanest possible application of Mertens' theorem: both factors converge absolutely (as they're the exponential series, with infinite radius of convergence), so the Cauchy product theory applies with no caveats — a sharp contrast to Problem 144's failure case where neither factor converges absolutely.
- The binomial identity $\sum_k\binom nk(-1)^k=0^n$ is a fundamental combinatorial fact, also underlying inclusion-exclusion arguments; its appearance here is a purely algebraic proof of $e\cdot e^{-1}=1$ that never explicitly evaluates either series as a transcendental number.
- This problem is the "safe" companion to Problems 137 and 144, both of which explore what goes wrong with Cauchy products absent absolute convergence — here, by contrast, everything works exactly as expected.
Result
The identity is proven: