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

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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?

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

1 Answer 1

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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.

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  • $\begingroup$ 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 $n>N$. .... $\endgroup$
    – Silent
    Commented Jan 16, 2019 at 3:11
  • $\begingroup$ 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? $\endgroup$
    – Silent
    Commented Jan 16, 2019 at 3:12
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    $\begingroup$ @Silent: Well done. Perhaps more directly in considering the limit as $x \to 1-$ we can assume $0 < x < 1$ and we have $(1-x)\sum_{n=N+1}^{m}\frac{\varepsilon}{2}|x^n| <(1-x)\sum_{n=N+1}^{\infty}\frac{\varepsilon}{2}|x^n| = (1-x)x^{N+1} \frac{\varepsilon}{2}\sum_{n=0}^\infty x^n = (1-x)x^{N+1}\frac{\varepsilon}{2}\frac{1}{1-x} < \frac{\varepsilon}{2}$ $\endgroup$
    – RRL
    Commented Jan 16, 2019 at 3:18
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    $\begingroup$ ... since $x^{N+1} < 1$ $\endgroup$
    – RRL
    Commented Jan 16, 2019 at 3:20
  • $\begingroup$ Thank you so much! $\endgroup$
    – Silent
    Commented Jan 16, 2019 at 3:29

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