I was given this homework
Suppose the radius of convergence of the series $\sum_{n=1}^\infty a_nx^n$ is $R$ and $a_n \geq 0$ for $n=1,2,\dots$. Prove that $\lim_{x \to R^-} \sum_{n=1}^\infty a_nx^n = \sum_{n=1}^\infty a_nR^n$.
Now the proof is extremely easy due to the assumption $a_n \geq 0$. If RHS is finite, then LHS=RHS by Abel’s theorem. If RHS is $+\infty$, then
\begin{align*} \lim_{x \to R^-} \sum_{n=1}^\infty a_nx^n > \lim_{x \to R^-} \sum_{n=1}^N a_nx^n + 0 \end{align*}
As we can pick $N$ large enough to make it large enough and we can plug in limit for finite $N$.
Why is the condition $a_n \geq 0$ necessary? I am thinking of doing
\begin{align*} \sum_{n=1}^\infty a_nx^n = \sum_{n=1}^N a_nR^n+ \sum_{n=1}^N a_n(x^n-R^n) + \sum_{n=N+1}^\infty a_nx^n \end{align*}
We can make the first term big. The second term can be small by sending $x$ sufficiently close to $R$. I think the problem comes from the third term.
Can someone give a concrete counter-example to why this doesn’t hold when we allow $a_n \in \mathbb{R}$ instead of $a_n \geq 0$?
Look at the power series $\displaystyle \sum_{k=0}^\infty (-1)^kx^k.$ The radius of convergence is $R = 1$ and for each $x \in (-1,1)$ the series sums to $\dfrac{1}{1+x}$. Is it true or false that $$\lim_{x \to 1^-} \frac 1{1+x} = \sum_{k=0}^\infty (-1)^k?$$