# $\lim\limits_{x\to\infty}f(x)^{1/x}$ where $f(x)=\sum\limits_{k=0}^{\infty}\cfrac{x^{a_k}}{a_k!}$.

Does the following limit exist? What is the value of it if it exists? $$\lim\limits_{x\to\infty}f(x)^{1/x}$$ where $f(x)=\sum\limits_{k=0}^{\infty}\cfrac{x^{a_k}}{a_k!}$ and $\{a_k\}\subset\mathbb{N}$ satisfies $a_k<a_{k+1},k=0,1,\cdots$

$\bf{EDIT:}$ I'll show that $f(x)^{1/x}$ is not necessarily monotonically increasing for $x>0$.

Since $\lim\limits_{x\to+\infty}\big(x+2\big)^{1/x}=1$, for any $M>0$, we can find some $L > M$ such that $\big(2+L\big)^{1/L}<\sqrt{3}$. It is easy to see that: $$\sum_{k=N}^\infty \frac{x^k}{k!} = \frac{e^{\theta x}}{N!}x^N\leq \frac{x^N}{N!}e^x,\quad \theta\in(0,1)$$ Hence we can choose $N$ big enough such that for any $x\in[0,L]$ $$\sum_{k=N}^\infty \frac{x^k}{k!} \leq 1$$ Now, we let $$a_k=\begin{cases}k,& k=0,1\\ 0,& 2\leq k <N\\ k,& k\geq N\end{cases}$$ Then $f(x)= 1+x+\sum\limits_{k=N}^\infty\frac{x^k}{k!}$ and $$f(2)^{1/2} \geq \sqrt{3} > (2+L)^{1/L} \geq f(L)^{1/L}$$ which shows that $f(x)^{1/x}$ is not monotonically increasing on $[2,L]$.

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${}{}{}{}{}{}{}$ – Pedro Tamaroff Nov 10 '12 at 16:21
I give +50 bounty – Leitingok Nov 14 '12 at 6:14

It does not have to. You know $e^x= \sum_0^\infty x^n/ n!$.

Now (using the convention that summing from or to a non-integer number means summing to or from their floor) $$\sum_0^{\sqrt{x}-1} x^n/ n! < \sum_0^{\sqrt{x}-1} x^n< x^{\sqrt{x}}$$ for large $x$,

and

$$\sum_{x^2}^{\infty} x^n/ n! < \sum_{x^2}^{\infty} x^n/ (x^2/e)^n< \sum_{x^2}^{\infty} (e/x)^n < 1$$ for large $x$.

So if we take the subsequence of $1,2,...$ consisting of those integers between all $2^{4k+2}$ and $2^{4k+4}$, but not the rest (that is from 5 to 16, then all from 65 to 256 etc) , then for $x=2^{4k+1}$ (like 2, 32 etc) we will get $\sum_0^\infty x^{a_k}/(a_k)! <x^{\sqrt{x}}+1$, so that $f(x)^{1/x}$ limits to $1$, and for $x=2^{4k+3}$ (like 8, 128 etc) we will get $\sum_0^\infty x^{a_k}/(a_k)! >e^x -x^{\sqrt{x}}-1$ so that $f(x)^{1/x}$ limits to $e$.

Edit: How to see the inequalities for $x=2^{2k+1}$ and $x=2^{2k+3}$, also known as "what is going on?"

The idea is that for each $x$ the sum $e^x=\sum x^n/n!$ is divided into 3 parts - $n$ from $1$ to $\sqrt{x}$, then $n$ from $\sqrt{x}$ to $x^2$ and then from $x^2$ to $\infty$, and the first and last parts are relatively small.

Now $\sum_0^\infty x^{a_k}/(a_k)!$ is the sum of only those $n$ that $a_k$ picks out form $\mathbb{N}$. So if we make a sequence that does not pick out anything from the middle part for some $x$, that is avoids the $n$ from $\sqrt{x}$ to $x^2$, then the sum for that $x$ would be smaller than the two remaining parts, that is $x^{\sqrt{x}}+1$.

Similarly, if the first and the last parts of the summ are small, then the middle part is big, at least $e^x-x^{\sqrt{x}}-1$. So if for some $x$ the subsequence hits all the integers in the middle part, then the sum will be at least that big for that $x$.

The sequence picking out the integers between $2^{4k+2}$ and $2^{4k+4}$ avoids the middle part for $x=2^{4k+1}$ and hits all of the middle part for $2^{4k+3}$. Hence the bounds, providing two subsequences where limit is 1 and $e$ correspondingly.

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For the first case I think you mean "$f(x)^{1/x}$ limits to 1", not to 0. – Lukas Geyer Nov 14 '12 at 23:21
Lukas - Thanks! – Max Nov 14 '12 at 23:57
@Max, how to get the inequalities when $x=2^{4k+1}$ and $x=2^{4k+3}$ – Leitingok Nov 16 '12 at 5:30
I have edit a comment explaining what is going on. – Max Nov 16 '12 at 7:29

This limit does not exist in general. First observe that for any polynomial $P$ with non-negative coefficients we have $$\lim_{x\to\infty} P(x)^{1/x} = 1$$ and $$\lim_{x\to\infty} (e^x - P(x))^{1/x} = \lim_{x\to\infty} e (1-e^{-x}P(x))^{1/x} = e.$$ For ease of notation let $$e_n(x) = \sum_{k=n}^\infty \frac{x^k}{k!} = e^x - \sum_{k=0}^{n-1} \frac{x^k}{k!}.$$ Note that $\lim\limits_{n\to\infty} e_n(x) = 0$ for every fixed $x$.

Now define a power series of the form $$f(x) = \sum_{i=1}^\infty \sum_{k=m_i}^{n_i} \frac{x^k}{k!},$$ along with partial sums $$P_j(x) = \sum_{i=1}^j \sum_{k=m_i}^{n_i} \frac{x^k}{k!},$$ where $1 \le m_1 \le n_1 < m_2 \le n_2 < \ldots$ are chosen inductively below. We want to find increasing sequences $(x_i)$ and $(y_i)$ with $x_i \to \infty$, $y_i \to \infty$, and $f(x_i)^{1/x_i} \le \frac32$ and $f(y_i)^{1/y_i} \ge 2$, which obviously implies non-existence of $\lim\limits_{x\to\infty} f(x)^{1/x}$.

Having already defined $m_i$, $n_i$, $x_i$, $y_i$ for $i < j$, we know that $\lim\limits_{x\to\infty} P_{j-1}(x)^{1/x} = 1$, so there exists $x_{j}>j+x_{j-1}$ such that $P_{j-1}(x_j)^{1/x_j} \le \frac54$. Then there exists $m_{j}>n_{j-1}$ such that $$(P_{j-1}(x_j)+e_{m_{j}}(x_{j}))^{1/x_j} \le \frac32,$$ which implies that whatever choices we make for $n_j$, $m_{j+1}$, etc., we always get $$f(x_j)^{1/x_j} \le (P_{j-1}(x_j)+e_{m_{j}}(x_{j}))^{1/x_j} \le \frac32.$$ We also know that $$\lim\limits_{x\to\infty} (P_{j-1}(x) + e_{m_j}(x))^{1/x} = e>2,$$ so there exists $y_j > x_j$ with $$(P_{j-1}(y_j) + e_{m_j}(y_j))^{1/y_j} >2.$$ Furthermore, there exists $n_j > m_j$ with $$(P_{j-1}(y_j) + e_{m_j}(y_j)- e_{n_j+1}(y_j))^{1/y_j} >2.$$ Lastly, this implies $$f(y_j)^{1/y_j} \ge P_j (y_j)^{1/y_j} = (P_{j-1}(y_j) + e_{m_j}(y_j)- e_{n_j+1}(y_j))^{1/y_j} >2.$$ By pushing this idea a little further, one can achieve $\liminf\limits_{x\to\infty} f(x)^{1/x} = 1$ and $\limsup\limits_{x\to\infty} f(x)^{1/x} = e$.

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