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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.

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  • $\begingroup$ Shouldn't the inequality in your conclusion for positive $a_n$ be the other way round? $\endgroup$ – maxmilgram Jan 24 at 10:04
  • $\begingroup$ You are searching for an Abel summable sequence. $\endgroup$ – Daniele Tampieri Jan 24 at 10:41
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    $\begingroup$ @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. $\endgroup$ – Ingix Jan 24 at 12:25
  • $\begingroup$ Writing $$\lim_{x \to 1} \sum_{n = 0}^\infty a_nx^n \ne \infty$$ is a little strange. Why not ask: Does $$\lim_{x \to 1^-} \sum_{n = 0}^\infty a_nx^n =\infty?$$ $\endgroup$ – zhw. Jan 24 at 21:52
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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 $\lim_{x \uparrow 1} f(x) = \infty$.

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  • $\begingroup$ How can you assume there is a finite a_i that a_i<0? $\endgroup$ – Shaq Jan 24 at 18:44
  • $\begingroup$ 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 $n \in \mathbb{N}$. $\endgroup$ – p4sch Jan 24 at 20:06
  • $\begingroup$ You are assuming $\lim_{\uparrow 1}f(x)$ exists? $\endgroup$ – zhw. Jan 24 at 21:46
  • $\begingroup$ 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. $\endgroup$ – p4sch Jan 25 at 10:18
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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.

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  • $\begingroup$ You mean $A_n = \sum_{ k = 0}^n \mid a_k \mid$, no ? $\endgroup$ – Thinking Jan 24 at 21:58
  • $\begingroup$ @Thinking Yes, thanks. Will correct. $\endgroup$ – zhw. Jan 24 at 22:00
  • $\begingroup$ 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 $a_n$. $\endgroup$ – Thinking Jan 24 at 22:17
  • $\begingroup$ @Thinking I certainly corrected the typo. It seems you have another question. I'll be back. $\endgroup$ – zhw. Jan 24 at 22:22
  • $\begingroup$ ? So $A_n$ is just $(n+1)\mid a_n \mid$ ? $\endgroup$ – Thinking Jan 24 at 22:23

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