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Show that $\sum_{k=1}^\infty \frac{1}{k!}$ converges without the Ratio Test.
(hint: show first that $k!\geq 2^{k-1}|\forall k \in \mathbb{N}$)

\begin{align} & k!\geq 2^{k-1} \\ \leftrightarrow\quad & k!\geq 2^k\cdot2^{-1} \\ \leftrightarrow\quad & 2\cdot k!\geq 2^k \\ \text{For }k=1:\quad& 2\cdot1!=2^1 \\ \text{For }k=2:\quad& 2\cdot2!=2^2 \\ \text{For }k=3:\quad& 2\cdot3!=2^3 \\ \leftrightarrow \quad& 12 \geq 8 \\ \text{For }k+1:\quad & 2(k+1)!=2(k+1)k! \geq 2^{k+1}=2\cdot2^k \\ \leftrightarrow \quad & (k+1)k! \geq 2^k \end{align}

that's true because we assume that $k!\geq 2^{k-1}\mid \forall k \in \mathbb{N}$

So it is proved by induction that $k!\geq 2^{k-1}|\forall k \in \mathbb{N}$ is true.

$s_n=\sum_{k=1}^\infty \frac{1}{k!}=1+1/2+1/6+1/24+\cdots$
if $k!\geq 2^{k-1}|\forall k \in \mathbb{N}$) then $(k!)^{-1}\leq 2^{1-k}$
$\leftrightarrow \sum_{k=1}^\infty \frac{1}{k!} \leq \sum_{k=1}^\infty 2^{1-k}$ with $\sum_{k=1}^\infty 2^{1-k}$ is a geometric serie, which converges
$\leftrightarrow |\frac{1}{k!}| \leq 2^{1-k}$ by the comparison test:
and $s_n$ converges.
Is that prove correct? Or should I try to prove it with the connection to e?Or is there a faster way to prove it?

  • 0
    Another way is to use comparison to $1/k^2$, or to $1/(k(k-1))$ (taking some care about the $k=1$ term).2012-12-21

3 Answers 3

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The complex exponential is entire and is defined by $f(z)=e^z=\sum_{k=0}^{\infty} \frac{z^k}{k!} \ .$ The series converges everywhere. For $z=1$ we have $e=1+\sum_{k=1}^{\infty} \frac{1}{k!}$

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    One usually uses the ratio test, but since you cannot use it, one possiblility is to use the root test and the the following link http://math.stackexchange.com/questions/136626/lim-limits-n-to-infty-sqrtnn-is-infinite2012-12-21
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You can also use root test. $\lim\limits_{n\to\infty}\sqrt[n] \frac{1}{n!}=0$ Since expression above is geometric mean of the sequence $\left(\frac{1}{n}\right)$, it converges to the limit of the sequence $\left(\frac{1}{n}\right)$. Then by root test, the series is convergent.

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Before I read your detailed answer, I posted:

The $n$th term lies between $0$ and ${2^{-n}}$ for $n>3$, and you already know that $\sum {2^{-n}}$ converges.

That's essentially the same as your answer. You asked if there was a shorter way to do it. I don't know if this counts as a shorter way to do it, but it's certainly a shorter way to say it. Don't you have a theorem you can appeal to that says that if $|a_i|\le |b_i|$ for all sufficiently large $i$, and if $\sum b_i$ converges, then so does $\sum a_i$?