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What is the answer to the following limit of a power series?

$$\lim_{x\rightarrow +\infty} \sum_{k=1}^\infty (-1)^k \left(\frac{x}{k} \right)^k$$

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    $\begingroup$ Sometime ($\approx 2.5$ years) back user deoxygerbe and myself noted some properties of the function (without the limit), which you can find here. The limit you are after is $$\lim_{x \to \infty} \left(\Upsilon(-x)-1 \right)$$ $\endgroup$ – user17762 May 25 '13 at 4:22
  • $\begingroup$ could you please explain a little bit more? Is it $\gamma$ function? $\endgroup$ – Mamal May 25 '13 at 4:26
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    $\begingroup$ @user78897 Added an answer. The function $\Upsilon(z)$ is defined in the article, I have linked to. $\endgroup$ – user17762 May 25 '13 at 4:31
  • $\begingroup$ @user17762 .... thank you so much... do you think the answer is bounded? $\endgroup$ – Mamal May 25 '13 at 4:50
  • $\begingroup$ @user17762 : neat idea .... Thank you $\endgroup$ – Mamal May 26 '13 at 3:39
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A simple calculation shows that

\begin{align*} \sum_{k=1}^{\infty} (-1)^{k} \left( \frac{x}{k} \right)^{k} &= \sum_{k=1}^{\infty} \frac{(-1)^{k} x^{k}}{(k-1)!} \int_{0}^{\infty} t^{k-1} e^{-kt} \, dt = \int_{0}^{\infty} \sum_{k=1}^{\infty} \frac{(-1)^{k} x^{k} t^{k-1} e^{-kt}}{(k-1)!} \, dt \\ &= -x \int_{0}^{\infty} \exp \left\{ - t \left( 1 + x e^{-t} \right) \right\} \, dt = - \int_{0}^{1} x \cdot u^{x u} \, du, \end{align*}

where $u = e^{-t}$. Now we claim that

$$ \lim_{x\to\infty} \int_{0}^{1} x \cdot u^{x u} \, du = 1. $$

To find the limit, we prove the following lemma:

Lemma. Let $f : [0, \delta] \to [0, 1]$ be a measurable function. Suppose there exists $0 < A < B$ such that $$ 1 - Ax \leq f(x) \leq 1 - Bx. $$ Then we have $$ \frac{1}{A} \leq \liminf_{x\to\infty} \left( x \int_{0}^{\delta} f(t)^{x} \, dt \right) \leq \limsup_{x\to\infty} \left( x \int_{0}^{\delta} f(t)^{x} \, dt \right) \leq \frac{1}{B}. $$

Assume this lemma holds. Let $f(u) = u^{u}$. Then we observe that

  1. $f(u)$ decreases for $[0, 1/e]$ and increases for $[1/e, 1]$.
  2. For any small $\epsilon > 0$, there exists small $\delta > 0$ such that $$ 1 - (1+\epsilon)(1-u) \leq f(u) \leq 1 - (1-\epsilon)(1-u) $$ for $0 < u < \delta$.
  3. For any large $M > 0$, we can choose small $\delta > 0$ such that $$f'(u) = u^{u}(1 + \log u) \leq -M$$ for $0 < x < \delta$. In particular, $f(u) \leq 1 - Mu$.

Let $\epsilon > 0$ be small and $M > 0$ be large. Let $\delta > 0$ be a sufficiently small number satisfying the conditions simultaneously. Then we have

$$ 0 \leq x \int_{\delta}^{1-\delta} u^{xu} \, du \leq x \max\{ f(\delta)^{x}, f(1-\delta)^{x} \} \xrightarrow{x\to\infty} 0. $$

Also, Lemma shows that

$$ \frac{1}{1+\epsilon} \leq \liminf_{x\to\infty} \left( x \int_{1-\delta}^{1} u^{xu} \, du \right) \leq \limsup_{x\to\infty} \left( x \int_{1-\delta}^{1} u^{xu} \, du \right) \leq \frac{1}{1-\epsilon} $$

and

$$ 0 \leq \limsup_{x\to\infty} \left( x \int_{0}^{\delta} u^{xu} \, du \right) \leq \frac{1}{M}. $$

Putting together, we have

$$ \frac{1}{1+\epsilon} \leq \liminf_{x\to\infty} \left( x \int_{0}^{1} u^{xu} \, du \right) \leq \limsup_{x\to\infty} \left( x \int_{0}^{1} u^{xu} \, du \right) \leq \frac{1}{1-\epsilon} + \frac{1}{M}. $$

Therefore, letting $M \to \infty$ and $\epsilon \to 0^{+}$, we obtain

$$ \lim_{x\to\infty} \left( x \int_{0}^{1} u^{xu} \, du \right) = 1 $$

as desired.

Proof of Lemma. For any $0 < \eta < \delta$, we have $$ 0 \leq x \int_{\eta}^{\delta} f(t)^{x} \, dt \leq x \int_{\eta}^{\delta} \max \{ 1- B\eta, 0 \}^{x} \, dt \leq \max \{ x \delta (1- B\eta)^{x}, 0 \} \xrightarrow[]{x\to\infty} 0. $$ Thus we may assume that $\delta$ is sufficiently small so that $1 - A\delta \geq 0$. Then \begin{align*} x \int_{0}^{1/A} (1 - At)^{x} \, dt + o(1) &= x \int_{0}^{\delta} (1 - At)^{x} \, dt \\ &\leq x \int_{0}^{\delta} f(t)^{x} \, dt \\ &\leq x \int_{0}^{\delta} (1 - Bt)^{x} \, dt = \leq x \int_{0}^{1/B} (1 - Bt)^{x} \, dt + o(1). \end{align*} Evaluating, we obtain $$ \frac{x}{A(x+1)} + o(1) \leq x \int_{0}^{\delta} f(t)^{x} \, dt \leq \frac{x}{B(x+1)} + o(1), $$ proving the lemma.

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  • $\begingroup$ Although one is inclined to believe it with exponentials, you might want to justify how you exchange the integral and the limit. $\endgroup$ – Martin Argerami May 25 '13 at 14:34
  • $\begingroup$ @MartinArgerami, I admit that in this problem, interchanging the limit is almost impossible to justify. My argument was to use the dominated convergence theorem, but my choice of upper bound was completely wrong. Indeed, it seems that no such dominating function exists. I will fix this issue as possible as I can. $\endgroup$ – Sangchul Lee May 25 '13 at 15:13
  • $\begingroup$ @sos440 Thats neat. Thank you so much... $\endgroup$ – Mamal May 26 '13 at 3:38
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It is straightforward to reproduce the integral representations of the sum found by @deoxygerbe and @user17762 in their interesting paper and by @sos440, $$\begin{eqnarray*} \sum_{k=1}^\infty (-1)^k \left(\frac{x}{k} \right)^k &=& -x\int_0^1 t^{x t} dt \\ &=& -x \int_0^\infty e^{-z(xe^{-z}+1)} dz. \end{eqnarray*}$$ Here we verify @sos440's result, that the sum is $-1$, by examining the last integral.

We have $$\begin{eqnarray*} x \int_0^\infty e^{-z(xe^{-z}+1)} dz &=& \underbrace{\int_0^1 f(z)dz}_{I_1} + \underbrace{\int_1^\infty f(z)dz}_{I_2}, \end{eqnarray*}$$ where $f(z) = x e^{-z(xe^{-z}+1)}$. On $(1,\infty)$, $f(z)$ has a global max at $z_0$, the solution to $x(z_0-1)=e^{z_0}$. Note that $z_0 \sim \log x + \log\log x$ so that $\lim_{x\to\infty} z_0 = \infty$. Applying Laplace's method we find $$\begin{eqnarray*} I_2 &\sim& \sqrt{\frac{2\pi}{(z_0-2)(z_0-1)}} e^{-z_0/(z_0-1)} \\ &\sim& \frac{\sqrt{2\pi}}{e z_0}. \end{eqnarray*}$$ Thus, in the limit, $I_2 = 0$.

But $$\begin{array}{ccccc} \displaystyle x\int_0^1 e^{-z(x+1)} dz &\le& \displaystyle x \int_0^1 e^{-z(xe^{-z}+1)} dz &\le& \displaystyle x\int_0^1 e^{-z[x(1-z)+1]} dz. \end{array}$$ (Here we use that on $[0,1]$, $1-z\le e^{-z}\le 1$.) The bounding integrals can be calculated explicitly and in the limit they are unity. Thus, in the limit, $I_1 = 1$.

Therefore, $$\begin{eqnarray*} \lim_{x\to\infty} \sum_{k=1}^\infty (-1)^k \left(\frac{x}{k} \right)^k &=& -\lim_{x\to\infty} x \int_0^\infty e^{-z(xe^{-z}+1)} dz \\ &=& -\lim_{x\to\infty} (I_1+I_2) \\ &=& -1. \end{eqnarray*}$$

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  • $\begingroup$ Nice... Thank you so much. $\endgroup$ – Mamal May 26 '13 at 3:38
  • $\begingroup$ @Mamal: Glad to help. Thank you for the interesting question. $\endgroup$ – user26872 May 26 '13 at 3:43
  • $\begingroup$ By the way, can we use the same way to find the limit of $\lim_{x \rightarrow +\infty} \sum_{k=1}^\infty(-1)^k(\frac{x}{k})^k k^m$? (for a fininite value of $m$) $\endgroup$ – Mamal May 26 '13 at 5:27
  • $\begingroup$ @Mamal: This looks like a harder problem. The first thing to notice is that if $D = x \partial/\partial x$, then $\sum_{k=1}^\infty (-1)^k (x/k)^k k^m = D^m \sum_{k=1}^\infty (-1)^k (x/k)^k$. You may want to open another question. $\endgroup$ – user26872 May 26 '13 at 5:57

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