Is this following proof correct?

The limit \begin{align*} \lim_{n\to\infty}\frac{t^{n+1}}{(n+1)!} \end{align*} tends to zero if for every $\epsilon>0$, there exists $N_\epsilon$ such that \begin{align*} n\geq N_\epsilon\implies \left|\frac{t^{n+1}}{(n+1)!}\right|<\epsilon \end{align*} For any $t\in\mathbb{R}$, there exists $N\in\mathbb{N}$ such that $t<N$. Let $N$ be the least such natural number. Then
\begin{align*} \frac{t^{N+k}}{(N+k)!}&=\frac{t\cdot t\cdots t\cdot t\cdots t\cdot t}{1\cdot 2\cdots (N-1)N\cdots(N+k-1)(N+k)}\\ &<\frac{t\cdot t\cdots t \cdot t\cdots t\cdot t}{1\cdot 1\cdots 1\cdot N\cdots N\cdot N}\\ &=t^N\left(\frac{t}{N}\right)^{k} \end{align*} Let $a=\frac{t}{N}$. Since $t<N$, $a=\frac{t}{N}<1$. If \begin{align*} \frac{t^{N+k}}{(N+k)!}<t^N\left(\frac{t}{N}\right)^{k}=t^Na^{k}<\epsilon \end{align*} then dividing both sides by $t^N$, \begin{align*} a^{k}<\frac{\epsilon}{t^N} \end{align*} Taking $\log_a$ of both sides reverses the inequality because it is a decreasing function. Hence \begin{align*} k>\log_{a}\left(\frac{\epsilon}{t^N}\right). \end{align*} Any value for $k$ greater than this expression will ensure the limit remains within the desired bounds. Therefore, take $N_\epsilon=N+k$. Since the choice of $t$ was arbitrary, we may conclude that the limit converges throughout the entirety of $\mathbb{R}$.


The proof looks good (actually very good) to me. The only improvement I can suggest is to turn it around a little, so that no cleverness is invoked.

Proof: Given $t>0$ and $\epsilon > 0$ arbitrary, put: $$\begin{align*} N &= \lfloor t \rfloor + 1 & (\text{$N$ is the smallest natural number strictly greater than 1}) \\ a &= \tfrac{t}{N} & (\text{so $a < 1$}) \\ k &= \left\lceil \log_{a}\left( \frac{\epsilon}{t^N} \right) \right\rceil \\ N_{\epsilon} &= N+k \end{align*}$$ Then if $n > N_{\epsilon}$ we have $$\begin{align*} \frac{t^{n+1}}{(n+1)!} &= \frac{t^{N+n-N+1}}{N!(N+1)(N+2)\dots(n+1)} \\ &< \frac{t^N}{1^N} \cdot \left( \frac{t}{N} \right)^{n+1-N} \\ &= t^N a^{n - N_{\epsilon} + N_{\epsilon} +1-N} \\ &< t^N a^{k+1} & (\text{via $N_{\epsilon} = N+k$, $n > N_{\epsilon}, a < 1$)} \\ &< t^N \frac{\epsilon}{t^N} & (\text{via construction of $k$}) \\ &= \epsilon \end{align*}$$ as required.

Remark: Your calculations did all the hard work of finding $N_{\epsilon}$. All this answer suggests is to summarize it into easier(?) steps from 'if $n > N_{\epsilon}$' to 'then $(\ldots) < \epsilon$'.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.