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proof of Abel's limit theorem
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(Proof)
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Without loss of generality we may assume $r=1$ , because otherwise we can set $a^\prime_n:=a_r^n$ , so that $\sum a^\prime_nx^n$ has radius $1$ and $\sum a^\prime$ is convergent if and only if $\sum a_nr^n$ is. We now have to show that the function $f(x)$ generated by $\sum a_nx^n$ (with $r=1$ )is continuous from below at $x=1$ if it is defined there. Let $s:=\sum a_n$ . We have to show that $$\lim_{x\to1^-}f(x)=s.$$ If $|x|<1$ we have:
with $s_n:=\sum_{i=0}^na_i$ . Now, since $s-s_n\to0$ as $n\to\infty$ we can choose an $N$ for every $\varepsilon>0$ such that $|s-s_n|<\frac{\varepsilon}{2}$ for all $m>N$ . So for every $0<x<1$ we have:
This is smaller than $\varepsilon$ for all $x<1$ sufficiently close to $1$ , which proves $$\lim_{x\to r^-}\sum a_nx^n=\sum a_nr^n=\sum\lim_{x\to r^-}a_nx^n.$$
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"proof of Abel's limit theorem" is owned by mathwizard.
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Cross-references: continuous, generated by, function, convergent, radius, without loss of generality
This is version 2 of proof of Abel's limit theorem, born on 2004-02-16, modified 2004-02-16.
Object id is 5583, canonical name is ProofofAbelsLimitTheorem.
Accessed 3057 times total.
Classification:
| AMS MSC: | 40A30 (Sequences, series, summability :: Convergence and divergence of infinite limiting processes :: Convergence and divergence of series and sequences of functions) |
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Pending Errata and Addenda
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