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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?

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    The idea is fine, there are mistakes/typos. For example you want the reciprocals of $2^{k-1}$. If you are going to use Comparison Test, then all you need is $\frac{1}{k!}\le \frac{1}{2^{k-1}}$.2012-12-20
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    @Cocopuffs I corrected it to $2^{1-k}$.@André Nicolas Yes, I wanted to write it this way, but isn't $\frac{1}{2^{k-1}} = 2^{1-k}$,or do I have to write it in fraction form, does it make a difference in proving?2012-12-20
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    @André Nicolas so I don't even need to mention the serie-form?and I can skip this $\sum_{k=1}^\infty \frac{1}{k!} \leq \sum_{k=1}^\infty 2^{1-k}$?2012-12-20
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    If you are using Comparison Test, it is one line after inequality. Just say that it is standard fact that $\sum \frac{1}{2^n}$ converges. If Comparison has not been proved yet, look at partials sums of the factorials series. They are increasing, and by the inequality they are bounded above.2012-12-20
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    The logic is ok. I think this is one of the most elementary proofs, so probably what your teacher wanted you to come up with. The business with $e$ is later in the course.2012-12-20
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    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

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