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.


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 (note that we can suppose that c>0, because we can w.l.o.g. increase $a_1$), 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 $\lim_{x \uparrow 1} f(x) = \infty$.
