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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?

Michael Hardy
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phil
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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 $\frac{1}{k!}\le \frac{1}{2^{k-1}}$. – André Nicolas Dec 20 '12 at 19:44
  • @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? – phil Dec 20 '12 at 19:53
  • @André Nicolas so I don't even need to mention the serie-form?and I can skip this $\sum_{k=1}^\infty \frac{1}{k!} \leq \sum_{k=1}^\infty 2^{1-k}$? – phil Dec 20 '12 at 20:09
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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. – André Nicolas Dec 20 '12 at 20:17
  • 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. – Jyrki Lahtonen Dec 20 '12 at 21:39
  • Another way is to use comparison to $1/k^2$, or to $1/(k(k-1))$ (taking some care about the $k=1$ term). – Gerry Myerson Dec 21 '12 at 01:22

3 Answers3

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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 $$e=1+\sum_{k=1}^{\infty} \frac{1}{k!}$$

PAD
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    If OP had the fact that the exponential series converges everywhere as a given, do you think OP would be asking the question? – anon Dec 20 '12 at 21:34
  • I dont know what he has in his mind. But he is asking at the end for a connection with $e$. – PAD Dec 20 '12 at 21:37
  • No, it's ok.I thought perhaps one could prove it over $\sum_{k=1}^\infty \frac{1}{k!}=e=\lim\limits_{n\rightarrow\infty} (1- \frac {1}{n})^n$ but it's nearly the same.(Anyway I believe I would have to prove the convergence of the sum) – phil Dec 21 '12 at 06:01
  • 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-infinite – PAD Dec 21 '12 at 12:22
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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.

guest
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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 $\sum a_i$?

MJD
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