value of $ \sum_{k=1}^{\infty} \frac{1}{k!}$

Question

Is $ \sum_{k=1}^{\infty} \frac{1}{k!}$ diverges?

First I check that \begin{align} \lim_{k\rightarrow\infty} \frac{1}{k!} =0 \end{align}

It means at some point the above summation add $0$, thus it seems to me the above series sum converges. But how do we get the value of that summation?

Do you know the Euler number $e$? – b00n heT Oct 16 '16 at 06:44 — b00n heT, Oct 16 '16 at 06:44
Do you know the Maclaurin series for the function $e^x$? – Brian M. Scott Oct 16 '16 at 06:47 — Brian M. Scott, Oct 16 '16 at 06:47
@b00nheT,@Brian M.Scoot, Okay i see. Thanks – phy_math Oct 16 '16 at 06:48 — phy_math, Oct 16 '16 at 06:48

score 5 · Accepted Answer · answered Oct 16 '16 at 06:48

First, note that if $\sum_{k=1}^{\infty} a_k$ is a series, then checking whether the terms $a_k$ tend to zero gives you only a necessary condition for convergence, not a sufficient one. Your series passed this test, but it doesn't necessarily mean that it converges.

To prove that it does converge, you can use various convergence tests that are available for positive series. For example, the ratios

$$ \frac{a_k}{a_{k-1}} = \frac{\frac{1}{k!}}{\frac{1}{(k-1)!}} = \frac{(k-1)!}{k!} = \frac{1}{k} \xrightarrow[k \to \infty]{} 0 $$

also tend to zero and so ratio test, the series converges.

The limit of this series can be shown to be $e - 1$ by using the Taylor series expansion of $e^x$ which is

$$ e^x := \sum_{k=0}^{\infty} \frac{x^k}{k!} $$

(and indeed converges to $e^x$) and plugging in $x = 1$.

value of $ \sum_{k=1}^{\infty} \frac{1}{k!}$

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