A power series $\sum_{n = 0}^\infty a_nx^n$ such that $\sum_{n=0}^\infty a_n= +\infty$ but $\lim_{x \to 1} \sum_{n = 0}^\infty a_nx^n \ne \infty$

Let's consider the power series $$\sum_{n = 0}^{\infty} a_nx^n$$ with radius of convergence $$1$$. Moreover let's suppose that : $$\sum_{n = 0}^{\infty} a_n= +\infty$$. Then I would like to find a sequence $$(a_n)_{n \in \mathbb{N}} \subseteq \mathbb{R}^{\mathbb{N}}$$ that respect the above condition and such that :

$$\lim_{x \to 1, x < 1} \sum_{n = 0}^\infty a_nx^n \ne +\infty$$

First I've noticed that $$a_n$$ can't be a positive sequence, since if it was the case we would have for all $$N$$ :

$$\lim_{x \to 1} \sum_{n = 0}^\infty a_nx^n \geq \sum_{n = 0}^N a_n$$

Hence we need some of the $$a_n$$ to be negative. Moreover I need to use the assumption that the sum at $$x = 1$$ diverges, because if the sum at $$x = 1$$ converges then Abel's theorem says that the limit at $$x \to 1$$ and the sum of the power series at $$x = 1$$ are equal.

Thank you.

• Shouldn't the inequality in your conclusion for positive $a_n$ be the other way round? – maxmilgram Jan 24 at 10:04
• You are searching for an Abel summable sequence. – Daniele Tampieri Jan 24 at 10:41
• @maxmilgram No, though it takes a moment to realize that. You have $\lim_{x\to 1}\sum_{n=0}^N a_nx^n=\sum_{n=0}^Na_n$ (just polynomials). This means $\sum_{n=0}^N a_nx^n\ge\sum_{n=0}^Na_n - \epsilon$ for all $x>1-\delta$ for some $\delta$, corresponding to a chose $\epsilon > 0$. With all $a_n$ positive, we get $\sum_{n=0}^\infty a_nx^n\ge\sum_{n=0}^Na_n - \epsilon$ for all $x>1-\delta$ for that $\delta$. From that follows what's in the post, because $\epsilon > 0$ can be chosen freely. – Ingix Jan 24 at 12:25
• Writing $$\lim_{x \to 1} \sum_{n = 0}^\infty a_nx^n \ne \infty$$ is a little strange. Why not ask: Does $$\lim_{x \to 1^-} \sum_{n = 0}^\infty a_nx^n =\infty?$$ – zhw. Jan 24 at 21:52

Such sequences don't exist. We prove this by contradiction:

Assume that $$A_n := \sum_{k=1}^n a_k \rightarrow \infty$$, $$f(x):=\sum_{k=1}^\infty a_k z^k$$ is convergent in the unit disc and $$\lim_{x \uparrow 1} f(x)$$ exists. Removing finite many $$a_i$$ doesn't change the behaviour of $$A_n$$ and also not the existence of the limes for $$x \uparrow 1$$. Thus, we may assume that $$A_n \ge 0$$ for all $$n \ge 1$$. Now use Abel summation in the form $$\label{1}\tag{1}\sum_{k=1}^n a_k x^k = A_n x^n - \sum_{k=1}^{n-1} A_k x^k (1 - x).$$ Because $$|A_n x^n| \le \sum_{j=1}^n |a_j| |x|^j$$ and the series is absolute convergent (as a power series) for $$|x| < 1$$, we see that $$|A_n x^n|$$ is bounded for fixed $$|x| < 1$$. Since $$|A_n x^n| \le C(x)$$, we get for all $$|y| < |x|$$ that $$|A_n y^n| \le C(x) |y/x|^n \rightarrow 0$$. Hence $$\lim_{n \rightarrow \infty } A_n y^n =0$$ for all $$|y| < |x|$$. Since $$x$$ was arbitary, we get this statement for all $$|y| <1$$ (not necessarily uniform convergence).

Letting $$n \rightarrow \infty$$ in \eqref{1} gives for $$|x| <1$$ that $$\label{2}\tag{2}f(x)=\sum_{k=1}^\infty a_k x^k = (1-x) \sum_{k=1}^\infty A_k x^k.$$ Since $$\lim_{x \uparrow 1} f(x):=c$$ exists, we have $$\sum_{k=1}^\infty A_k x^k \sim \frac{1}{1-x}.$$ The Hardy–Littlewood tauberian theorem already implies that $$\sum_{k=1}^n (n+1-k) a_k = \sum_{k=1}^n A_k \sim n.$$ But, we have $$\frac{1}{2n}\sum_{k=1}^{2n} A_k \ge \frac{1}{2} A_n \rightarrow \infty.$$ A contradiction!

The problem changes rapidly, if we only require that $$\sum_{k=1}^n a_k$$ is not convergent.

For example take $$a_k = (-1)^k$$: We have $$\sum_{k=0}^\infty (-1)^k x^k = \frac{1}{x+1}$$ and that $$\lim_{x \uparrow 1} (x+1)^{-1} = 1/2$$, but $$\sum_{k=0}^n (-1)^k$$ is not convergent.

As zhw shows in the second answer, we can also prove that $$\lim_{x \uparrow 1} f(x) = \infty$$.

Note for this that the identity in \eqref{2} implies, since $$A_n >K$$ for all $$n \ge N$$ that $$f(x) \ge (1-x) K \sum_{k=N}^\infty x^k = K x^N$$ and thus $$\liminf_{x \uparrow 1} f(x) \ge K$$. Because $$K>0$$ is arbitary large, we get already $$\lim_{x \uparrow 1} f(x) = \infty$$.

• How can you assume there is a finite a_i that a_i<0? – Shaq Jan 24 at 18:44
• We don't know that $a_i <0$ only for finite many $i$, but we know that $A_n >0$ for all $n \ge N$ for some $N \in \mathbb{N}$, because $A_n \rightarrow \infty$. So removing all negative numbers of $a_1,\ldots,a_N$ gives $A_n \ge 0$ for all $n \in \mathbb{N}$. – p4sch Jan 24 at 20:06
• You are assuming $\lim_{\uparrow 1}f(x)$ exists? – zhw. Jan 24 at 21:46
• Yes, my argument shows that the limes doesn't exist. In fact, your argument shows that the sequence tend to $+\infty$. I have added also your argument to my answer. In fact, it is a more straight forward solution. – p4sch Jan 25 at 10:18

The following proof that $$\lim_{x\to 1^-}\sum_{n=0}^{\infty}a_nx^n = \infty$$ seems simpler to me..

Lemma: Let $$A_n=\sum_{k=0}^{n}|a_k|.$$ Then $$\sum_{n=0}^{\infty}A_nx^n<\infty$$ for $$x\in (0,1).$$

This follows from the fact that the radius of convergence is $$1,$$ which is the same as saying $$\limsup |a_n|^{1/n} =1.$$ This is a nice exercise. (A proof of the lemma is now in the comments.)

To prove the main result, let $$S_n=\sum_{k=0}^{n}a_n;$$ set $$S_{-1}=0.$$ Let $$x\in (0,1).$$ Then

$$\tag 1\sum_{n=0}^{\infty}a_nx^n = \sum_{n=0}^{\infty}(S_n-S_{n-1})x^n.$$

Now because $$|S_n| \le A_n,$$ the lemma shows we can write the last series as the difference of two convergent series, i.e. as

$$\sum_{n=0}^{\infty}S_nx^n - \sum_{n=1}^{\infty}S_{n-1}x^n = \sum_{n=0}^{\infty}S_nx^n - \sum_{n=0}^{\infty}S_{n}x^{n+1} = \sum_{n=0}^{\infty}S_nx^n(1-x).$$

Now let $$M>0.$$ Then there is $$N$$ such $$S_n>M$$ for $$n>N.$$ Write the last series as

$$(1-x)\left (\sum_{n=0}^{N}S_nx^n +\sum_{n=N+1}^{\infty}S_nx^n\right ) > (1-x)\left (\sum_{n=0}^{N}S_nx^n +Mx^{N+1}\frac{1}{1-x}\right ).$$

The $$\liminf_{x\to 1^-}$$ of the expression on the right equals $$0 + M =M.$$ We have thus shown the $$\liminf_{x\to 1^-}$$ of the left side of $$(1)$$ is $$\ge M.$$ Since $$M$$ was arbitrary, this $$\liminf$$ is $$\infty.$$ Thus the limit of left side of $$(1)$$ is $$\infty$$ as desired.

• You mean $A_n = \sum_{ k = 0}^n \mid a_k \mid$, no ? – Thinking Jan 24 at 21:58
• @Thinking Yes, thanks. Will correct. – zhw. Jan 24 at 22:00
• You didn't completely correct the typo. I know that if we denote $K_n = \sum_{k = 0}^n a_k$ then $\sum_{n = 0}^\infty K_n x^n$ converges since it comes from a Cauchy product. But I don't see why $\sum_{k = 0}^n A_kx^k$ converges, you don't know if the Cauchy test is going to work on the sequence $a_n$. – Thinking Jan 24 at 22:17
• @Thinking I certainly corrected the typo. It seems you have another question. I'll be back. – zhw. Jan 24 at 22:22
• ? So $A_n$ is just $(n+1)\mid a_n \mid$ ? – Thinking Jan 24 at 22:23