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Question Number 49961 by maxmathsup by imad last updated on 12/Dec/18

f i n d t h e s e q u e n c e (a_{n}) w i c h v e r i f y

(\sum_{n = 1}^{\infty} x^{n}) (\sum_{n = 0}^{\infty} \frac{{(- x)}^{n}}{n + 1}) = \sum_{n = 0}^{\infty} a_{n} x^{n} a l s o f i n d t h e r a d i u s o f t h i s s e r i e .

Commented by Abdo msup. last updated on 13/Dec/18

l e t α_{n} = 1 \forall n \in N a n d β_{n} = \frac{{(- 1)}^{n}}{n + 1} s o

(\sum_{n = 1}^{\infty} x^{n}) . (\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n + 1} x^{n}) = (\sum_{n = 0}^{\infty} α_{n} x^{n}) . (\sum_{n = 0}^{\infty} β_{n} x^{n})

\sum_{n = 0}^{\infty} a_{n} x^{n} / a_{n} = \sum_{i + j = n} α_{i} β_{j} = \sum_{i = 0}^{n} \frac{{(- 1)}^{n - i}}{n - i + 1}

\Rightarrow a_{n} = \sum_{p = 0}^{n} \frac{{(- 1)}^{p}}{p + 1} = \sum_{p = 1}^{n + 1} \frac{{(- 1)}^{p - 1}}{p}

w e h a v e R a d i u s (Σ x^{n}) = 1 a n d R a d i u d (Σ \frac{{(- x)}^{n}}{n + 1}) = 1 \Rightarrow

R e > a d i u s (Σ a_{n} x^{n}) ⩽ 1 .