Convergence of partial sums of sequence

Question

Suppose that $x\in[0,1]$. Let $S_k(x):=\sum_{n=1}^k \dfrac{(-1)^n}{n} x^n$, $S(x)=\sum_{n=1}^\infty \dfrac{(-1)^n}{n} x^n$. Prove that if $x_k\to1^-$, then $S_k(x_k)\to S(1)$ as $k\to\infty$.

I know that the series converges absolutely on $[0,1)$ and converges uniformely on $[0,1]$. What exactly do I have to do here? I am quite confused about the limit $S_k(x_k)$. If I had to compute $S(x_k)$ I know that I can interchange limit and summation for example if the series converges absolutely. But here the sum also depends on $k$.

Answer 1

You already know that $S_k \to S$ uniformly on $[0,1]$. Since each $S_k$ is continuous so is $S$. Now $|S_k(x_k)-S(1)| \leq |S_k(x_k)-S(x_k)|+|S(x_k)-S(1)|$. Given $\epsilon >0$ there exists $m$ such that $k>m$ implies $|S_k(x)-S(x)|<\epsilon /2$ for all $x$, in particular for $x=x_k$. So $|S_k(x_k)-S(x_k)| <\epsilon /2 $ for $k >m$. By continuity of $S$ there exists $j$ such that $|S(x_k)-S(1)|<\epsilon$ for $k >j$. If $k >\max \{m,j\}$ we get $|S_k(x_k)-S(1)| <\epsilon $. Hence $S_k(x_k) \to S(1)$.

Answer 2

You have $S_k(x_k) - S(1) = \left(S_k(x_k) - S(x_k)\right)+ \left(S(x_k)-S(1)\right)$. You mention in your question that you know that $\lim\limits_{k \to \infty} S(x_k)-S(1) = 0$.

So we're left to prove that

$$\lim\limits_{k \to \infty} S_k(x_k) - S(x_k) = \sum_{n=k+1}^\infty \dfrac{(-1)^n}{n}x_k^n = 0.$$

Denote $R_k = \displaystyle \sum_{n=k+1}^\infty \dfrac{(-1)^n}{n}$. Note two things:

1. $\lim\limits_{k \to \infty} R_k = 0$ as $\sum \dfrac{(-1)^n}{n}$ converges.
2. $\dfrac{(-1)^n}{n} = R_{n+1} - R_n$ for all $n \in \mathbb N$.

Therefore: $$S_k(x_k) - S(x_k) = \sum_{n=k+1}^\infty \left(R_{n+1} - R_n\right)x_k^n = -R_{k+1}x_k^{n+1}+(1-x_k)\sum_{n=k+1}^\infty R_{n+1} x_k^n. \tag{I}$$

Using a well know result of alternate series, you have $\vert R_k \vert \le \dfrac{1}{k}$ for all $k \in \mathbb N$. As $\vert x_k \vert \le 1$, the first term of $(I)$ converges to zero. The second can be bounded in absolute value by $$\vert 1-x_k \vert \sum_{n=k+1}^\infty \dfrac{x_k^n}{n} \le \vert 1- x_k \vert \vert \log(1-x_k)\vert.$$

The fact that the RHS of above inequality converges to $0$ as $k \to \infty$ completes the proof.