I'm already lost by the second line, as I don't understand where the final equality comes from. Specifically, I don't know how to get from the middle expression to the third sum.
Here is a picture of the problematic part of the proof.
What is going on in Theorem 8.2
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$\begingroup$
real-analysis
power-series
proof-explanation
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2Can you give us something more specific than "I'm lost" or "I don't understand"? Not sure how to answer your question if we don't know what your question is. – 2017-02-08
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0Yes, one second. – 2017-02-08
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0Ok @NickPeterson I updated it – 2017-02-08
1 Answers
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Note that $$ \begin{align*} \sum_{n=0}^{m}(s_n-s_{n-1})x^n&=\sum_{n=0}^{m}s_nx^n-\sum_{n=0}^{m}s_{n-1}x^n\\ &=\sum_{n=0}^{m}s_nx^n-\left[s_{-1}+x\sum_{n=1}^{m}s_{n-1}x^{n-1}\right]\\ &=\sum_{n=0}^{m}s_nx^n-x\sum_{n=0}^{m-1}s_nx^n\\ &=s_mx^m+\sum_{n=0}^{m-1}s_nx^n-x\sum_{n=0}^{m-1}s_nx^n\\ &=s_mx^m+(1-x)\sum_{n=0}^{m-1}s_nx^n \end{align*} $$
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1how do you $\LaTeX$ that fast? That's just unfair. – 2017-02-08
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1@Thomas Lived there for too long, I guess. :-) Every assignment/paper/research note for 9 years of my life – 2017-02-08
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0Thanks for the answer. I know it's a trivial question – 2017-02-08