# What is the answer to this limit

what is the limit value of the power series:

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

where $m>1$.

• It grows with out bound.... – Ethan May 27 '13 at 3:39
• – Mhenni Benghorbal May 27 '13 at 3:55
• @Benghorbal your mentioned problem is for $m=0$ – Mamal May 27 '13 at 4:00
• @Ethan... I dont think so... Do you have any proof for that? – Mamal May 27 '13 at 4:08
• @Mamal: By the way, what's the origin of this problem? – Mhenni Benghorbal May 27 '13 at 4:17

My solution is also similar. Let $\left\{ {a \atop b} \right\}$ be the Stirling number of the second kind, as Mhenni Benghorbal pointed out. Then from the identity

$$x^{n} = \sum_{j=0}^{n} \left\{ {n \atop j} \right\} (x)_{j},$$

where $(x)_{j} = x (x-1) \cdots (x-j+1)$ is the failing factorial, we obtain

$$x^{n} = \sum_{j=0}^{n} \left\{ {n+1 \atop j+1} \right\} (x-1)_{j}.$$

This allow us to write

\begin{align*} \frac{1}{k^{k-m}} & = \frac{k^{m}}{k^k} = \sum_{j=0}^{m} \left\{ {m+1 \atop j+1} \right\} \frac{1}{(k-1-j)!} \frac{\Gamma(k)}{k^{k}} \\ &= \sum_{j=0}^{m} \left\{ {m+1 \atop j+1} \right\} \frac{1}{(k-1-j)!} \int_{0}^{\infty} t^{k-1}e^{-kt} \, dt. \end{align*}

Here, we use the convention that $\frac{1}{j!}$ is zero when $j < 0$, which is compatible with the fact that the reciprocal of the Gamma function is an entire function with zeros at nonpositive integers. Then

\begin{align*} \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k^{k-m}} x^{k} &= \sum_{k=1}^{\infty} (-1)^{k-1} x^{k} \sum_{j=0}^{m} \left\{ {m+1 \atop j+1} \right\} \frac{1}{(k-1-j)!} \int_{0}^{\infty} t^{k-1}e^{-kt} \, dt \\ &= x \sum_{j=0}^{m} \left\{ {m+1 \atop j+1} \right\} \int_{0}^{\infty} \left( \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{(k-1-j)!} \left( x t e^{-t} \right)^{k-1} \right) \, e^{-t} \, dt \\ &= x \sum_{j=0}^{m} \left\{ {m+1 \atop j+1} \right\} \int_{0}^{\infty} \left( \sum_{k=0}^{\infty} \frac{(-1)^{k+j}}{k!} \left( x t e^{-t} \right)^{k+j} \right) \, e^{-t} \, dt \\ &= x \sum_{j=0}^{m} (-1)^{j} \left\{ {m+1 \atop j+1} \right\} \int_{0}^{\infty} \left( x t e^{-t} \right)^{j} \exp \left( x t e^{-t} \right) \, e^{-t} \, dt \\ &= \sum_{j=0}^{m} (-1)^{j} \left\{ {m+1 \atop j+1} \right\} x^{j+1} \int_{0}^{\infty} t^{j} e^{-tj} \exp \left( x t e^{-t} \right) \, e^{-t} \, dt. \end{align*}

Here, I considered the negation of your sum in order for typographical simplicity. Anyway, with the substitution $u = e^{-t}$, we obtain

\begin{align*} \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k^{k-m}} x^{k} &= \sum_{j=0}^{m} (-1)^{j} \left\{ {m+1 \atop j+1} \right\} x^{j+1} \int_{0}^{1} u^{xu} (-u \log u)^{j} \, du. \tag{1} \end{align*}

To analyze the behavior of

$$f_{j}(x) = x^{j+1} \int_{0}^{1} u^{xu} (-u \log u)^{j} \, du,$$

we introduce two functions $g_0 : [0, 1/e] \to [0, 1/e]$ and $g_1 : [0, 1/e] \to [1/e, 1]$ ad follows:

The graph of $y = -x \log x$ on $[0, 1]$ can be divided into two parts:

the increasing part and the decreasing part. Thus on each part, we can define the inverse function $g_0$ and $g_1$ as follows:

It is clear from the slope of the graph $y = - x \log x$ that $g_{0}'(0) = 0$ and $g_{1}' (0) = -1$. Also, obviously both $g_{0}'(y)$ and $g_{1}'(y)$ are integrable on $[0, 1/e]$. Keeping these in mind, we make the change of variable as follows:

\begin{align*} f_{j}(x) &= x^{j+1} \left[ \int_{0}^{1/e} e^{x u \log u} (-u \log u)^{j} \, du + \int_{1/e}^{1} e^{x u \log u} (-u \log u)^{j} \, du \right] \\ &= x^{j+1} \left[ \int_{0}^{1/e} y^{j} e^{-xy} g_{0}'(y) \, dy + \int_{0}^{1/e} y^{j} e^{-xy} (- g_{1}'(y) ) \, dy \right] \\ &= \int_{0}^{\infty} x^{j+1} y^{j} e^{-xy} \left( g_{0}'(y) \mathrm{1}_{[0, 1/e]}(y) \right) \, dy + \int_{0}^{\infty} x^{j+1} y^{j} e^{-xy} \left( - g_{1}'(y) \mathrm{1}_{[0, 1/e]}(y) \right) \, dy \end{align*}

Here, observe that the family of mappings

$$y \mapsto \frac{1}{j!} x^{j+1} y^{j} e^{-xy}$$

serves as an approximation to the identity as $x \to \infty$. This proves that

$$f_{j}(x) \xrightarrow{x\to\infty} j! (g_{0}'(0) - g_{1}'(0)) = j!.$$

So taking limit $x \to \infty$ to the identity $(1)$, we obtain

\begin{align*} \lim_{x\to\infty} \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k^{k-m}} x^{k} &= \sum_{j=0}^{m} (-1)^{j} \left\{ {m+1 \atop j+1} \right\} j!. \end{align*}

But it is known that the last sum is equal to

$$\sum_{j=0}^{m} (-1)^{j} \left\{ {m+1 \atop j+1} \right\} j! = \delta_{0m} = \begin{cases} 1 & m = 0 \\ 0 & m \geq 1 \end{cases}.$$

Therefore we have

$$\lim_{x\to\infty} \sum_{k=1}^{\infty} \frac{(-1)^{k}}{k^{k-m}} x^{k} = -\delta_{0m}.$$

The following graphs corresponds to $m = 1, \cdots, 5$ in zigzag order.

Added. Approximation to the identity is a way of justifying the concept of the Dirac delta. Consider a family of functions $\{ K_{t}(x) \}$ (parametrized by $t$ in this example) such that

$$\int_{\Bbb{R}} K_{t}(x) \,dx = q \quad \text{and} \quad \lim_{t\to\infty} \int_{|x|\geq \eta} K_{t}(x) \, dx = 0 \quad \text{for} \ \eta > 0.$$

That is, we cant interpret $K_{t}(x)$ as a mass density with the unit total mass such that the mass concentrates to the origin as $t \to \infty$. So it is natural to expect that $K_{t}(x)$ converges to the Dirac delta $\delta(x)$ in some sense as $t \to \infty$. The theory of approximation to the identity justifies this naive idea under various assumptions and in various mode of convergence.

In our problem,

$$K_{x}(y) = x f(xy) \quad \text{where} \quad f(y) = \frac{y^{j}}{j!}e^{-y} \quad \text{on} \ [0, \infty).$$

Thus we expect that as $x \to \infty$,

\begin{align*} f_{j}(x) &= j! \int_{0}^{\infty} \left\{ g_{0}'(y) - g_{1}'(y) \right\} \mathbf{1}_{[0, 1/e]}(y) K_{x}(y) \, dy \\ &= j! \int_{0}^{\infty} \left\{ g_{0}'(y) - g_{1}'(y) \right\} \mathbf{1}_{[0, 1/e]}(y) \delta(y) \, dy \\ &= j! \left\{ g_{0}'(0) - g_{1}'(0) \right\} = j!. \end{align*}

Indeed, some simple estimation proves this.

• Wow... Thats great... Thank you so much... – Mamal May 27 '13 at 7:04
• could you please explain a little bit more about the $y\rightarrow \frac{1}{j!}x^{j+1}y^{j}e^{-xy}$ – Mamal Jun 6 '13 at 6:12
• @Mamal, I can't grasp what you are questioning. Exactly what property of that function do you want to know? – Sangchul Lee Jun 7 '13 at 5:20
• Thank you very much, It was interesting. – Mamal Jun 8 '13 at 3:06

For $\lim_{x\rightarrow +\infty} \sum_{k=1}^\infty (-1)^k \frac{x^k}{k^{k-m}}$, the ratio of consecutive terms is

\begin{align} \frac{x^{k+1}}{(k+1)^{k+1-m}}\big/\frac{x^k}{k^{k-m}} &=\frac{x k^{k-m}}{(k+1)^{k+1-m}}\\ &=\frac{x }{k(1+1/k)^{k+1-m}}\\ &=\frac{x }{k(1+1/k)^{k+1}(1+1/k)^{-m}}\\ &\approx\frac{x }{ke(1+1/k)^{-m}}\\ &=\frac{x (1+1/k)^{m}}{ke}\\ \end{align}

For large $k$ and fixed $m$, $(1+1/k)^{m} \approx 1+m/k$, so the ratio is about $x/(ke)$, so the sum is like $\lim_{x\rightarrow +\infty} \sum_{k=1}^\infty (-1)^k \big(\frac{x}{ek}\big)^k$.

Note: I feel somewhat uncomfortable about this conclusion, but I will continue anyway.

Interesting how the $m$ goes away (unless I made a mistake, which is not unknown).

In the answer pointed to by Mhenni Benghorbal, it is shown that this limit is $-1$, so it seems that this is the limit of this sum also.

Here is a start. Using the integral representation derived in a previous problem and the identity

$$(xD)^m=\sum_{r=0}^{m}\begin{Bmatrix} m\\r \end{Bmatrix}x^r D^r,$$

where $D=\frac{d}{dx}$ and $\begin{Bmatrix} m\\r \end{Bmatrix}$ are Stirling numbers of the second kind.

$$\sum_{k=1}^\infty (-1)^k \frac{x^k}{k^{k}} = - \int_{0}^{1} x \cdot u^{x u} \, du .$$

$$\implies (xD)^m\sum_{k=1}^\infty (-1)^k \frac{x^k}{k^{k}}=\sum_{k=1}^\infty (-1)^k k^m\frac{x^k}{k^{k}}=-(xD)^m \int_{0}^{1} x \cdot u^{x u} \, du.$$

$$\implies \sum_{k=1}^\infty (-1)^k k^m\frac{x^k}{k^{k}}=\sum_{r=0}^{m}\begin{Bmatrix} m\\r \end{Bmatrix} \int_{0}^{1} f(x,u) du ,$$

where $f(x,u)$ is a function in $x$ and $u$, see the note. Now, it is your job to try analyzing these integrals and see what the limit is.

Notes:

1) We used the identity

$$D^r xe^{ax}= {a}^{r-1}{{\rm e}^{ax}} \left( ax+r \right) .$$

For finding the $n$th derivative, see section 6.

2)

$$f(x, u) = { {m{x}^{2\,m} \left( \ln \left( u \right)\right) ^{m-1}{u}^{ux+m -1}}{ }}+{x}^{2\,m+1} \left( u\ln \left( u \right) \right) ^{m}{u}^{ux}.$$

• Thank you... :)... However analyzing theseintegrals doesn't seem be easy... – Mamal May 27 '13 at 4:57
• @Mamal: What you are asking is not easy too!! By the way, they should not be that hard. Check asymptotic analysis of integrals. For instance, Laplace technique. I'll have a look at them when time allows. – Mhenni Benghorbal May 27 '13 at 4:58
• @Benghorbal: Yes, youre right. Thank you anyway, your answer seems to be a good start to the solution. – Mamal May 27 '13 at 5:01
• Using some simulations I figured that the answer is perhaps $0$, But Im not sure at all. – Mamal May 27 '13 at 5:03
• @Benghorbal: Thnk you so much... I'll be so thankful if you tell me your concolusion about the answer of the limit:) – Mamal May 27 '13 at 5:14