Solve $\sum_{k=0}^\infty\frac1{k!}$ without Taylor expansion or any expansion.

Question

Evaluate $\sum_{k=0}^\infty\frac1{k!}$ without Taylor, Mclauren or any expansion..

So I know that $$f(x)=e^x=\sum_{k=0}^\infty\frac{x^k}{k!}$$

around $0$ and for $x=1$ we have that: $\sum_{k=0}^\infty\frac1{k!}=e$, but however can you solve it without any expansion?

Answer 1

One can look up at the Proof Wiki. There is an arcticle, to be precise this one, which provides a proof of the equivalence of the limit and the series represantation of $e$.

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As achille hui pointed out the linked proof has got a little problem. As it is said at the bottom of the Proof Wiki page:

This article, or a section of it, needs explaining, namely:

Wait! That's invoking $\lim \sum=\sum\lim$ tacitly. Us sneaky bastards!

You can help Pr$\infty$fWiki by explaining it. To discuss this point in more detail, feel free to use the talk page. If you are able to explain it, then when you have done so you can remove this instance of {{Explain}} from the code.

Therefore the proof is not completely reliable.

Answer 2

You give us an infinite series and ask to evaluate it without any "expansion". This sounds a little contradictory.

As this rules out any alternative definition of $e$ (a transcendental number, which is hard to express with finite means), all we can do is

• prove convergence of the series (growing and bounded by $1+\sum_{k=0}^\infty2^{-k}=3$);

• evaluate approximations as tight as one wants by estimating the value of the tail.

E.g.,

$$1+1+\frac12+\frac1{3!}+\frac1{4!}<e<1+1+\frac12+\frac1{3!}+\frac1{4!}+\frac3{5!}$$

because

$$\frac{5!}{5!}+\frac{5!}{6!}+\frac{5!}{7!}+\cdots<\frac1{0!}+\frac1{1!}+\frac1{2!}+\cdots=e<3$$