# Power series continuous on endpoint if convergent there

This is Theorem 8.2 from baby Rudin, where it is assumed that series given has radius of convergence $$1$$. :

I can't understand last couple of lines in proof fully. In last line we have to derive $$(1-x)\sum_{n=0}^{N}|s_n-s||x|^n<\frac{\varepsilon}{2}$$ for some $$\delta>0$$ with $$x>1-\delta$$, and $$\left|(1-x)\sum_{n=N+1}^{\infty}(s_n-s)x^n\right|<\frac{\varepsilon}{2}.$$ Here, second inequality, I think, holds because, if $$\sum_{i=1}^\infty a_i$$ is convergent series, then $$\sum_{i=n}^\infty a_n\to0$$ as $$n\to0$$. How to get first inequality?

• Just for clarification, you are asking about how to get $(1-x)\sum_{n=0}^{N}|s_n-s||x|^n<\frac{\varepsilon}{2}$?
– ℋolo
Commented Jan 14, 2019 at 20:10
• @Holo Yes, I am sorry if I did not mention it clearly. Commented Jan 15, 2019 at 4:01

Since $$|x| < 1$$ it follows that

$$\sum_{n=0}^{N}|s_n-s||x|^n < \sum_{n=0}^{N}|s_n-s| = A$$

where $$A$$ is a constant for fixed $$N$$ independent of $$x$$.

Thus, with $$x < 1$$, we have as $$x \to 1-$$,

$$0 \leqslant (1-x)\sum_{n=0}^{N}|s_n-s||x|^n < A(1-x) \to 0$$

You are on the right track for the second inequality since $$|s_n - s| < \epsilon$$ for all sufficiently large $$n$$, and for $$0 < x < 1$$

$$\sum_{n=N+1}^{\infty}|x|^n = \frac{x^{N+1}}{1-x}$$

Try to finish from here.

• Thank you very much for this reply. For second inequality, I went like this: $$(1-x)\left|\sum_{n=N+1}^{m}(s_n-s)x^n\right|\le(1-x)\sum_{n=N+1}^{m}|(s_n-s)||x^n|\le(1-x)\sum_{n=N+1}^{m}\frac{\varepsilon}{2}|x^n|$$ for $N$ such that $|s_n-s|<\frac{\varepsilon}{2}$ for all $n>N$. .... Commented Jan 16, 2019 at 3:11
• Now, since $\sum_{n=N+1}^{\infty}|x|^n = \frac{x^{N+1}}{1-x}$, we see that $\sum_{i=n}^{\infty} x^n\to0$, so for some $N_1$ we see $\left|\sum_{i=n}^{\infty}x^n\right|<\frac{\varepsilon}{2}$ for all $n>N_1$. Now, for $N_0=\max\{N,N_1\}$, our desired inequality $$\left|(1-x)\sum_{n=N_0+1}^{\infty}(s_n-s)x^n\right|<\frac{\varepsilon}{2}$$ holds. Is this correct? Commented Jan 16, 2019 at 3:12
• @Silent: Well done. Perhaps more directly in considering the limit as $x \to 1-$ we can assume $0 < x < 1$ and we have $(1-x)\sum_{n=N+1}^{m}\frac{\varepsilon}{2}|x^n| <(1-x)\sum_{n=N+1}^{\infty}\frac{\varepsilon}{2}|x^n| = (1-x)x^{N+1} \frac{\varepsilon}{2}\sum_{n=0}^\infty x^n = (1-x)x^{N+1}\frac{\varepsilon}{2}\frac{1}{1-x} < \frac{\varepsilon}{2}$
– RRL
Commented Jan 16, 2019 at 3:18
• ... since $x^{N+1} < 1$
– RRL
Commented Jan 16, 2019 at 3:20
• Thank you so much! Commented Jan 16, 2019 at 3:29