- #1
Mathman23
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Hello
I have this problem here:
Given the series
[tex]\sum_{n=1} ^{\infty} \frac{x^{n+1}}{n(n+1)}[/tex]
show that this converges for every [tex]x \in \{ w \in \mathbb{C} \| \|w \| \leq 1 \}[/tex]
Solution:
Since
[tex] \sum_{n=1} ^{\infty} \left| \frac{x^{n+1}}{n(n+1)} \right| = \sum_{n=1} ^{\infty} \frac{1}{n(n+1)} \right| [/tex]
Since [tex]\sum_{n=1} ^{\infty} \frac{1}{n(n+1)} \right| [/tex] convergence then according the definition then
[tex]\sum_{n=1} ^{\infty} \frac{x^{n+1}}{n(n+1)}[/tex] is absolute convergent.
Thus
[tex]x \in \{ w \in \mathbb{C} \| \|w \| \leq 1 \}[/tex]. Right?
Best Regards
Fred
I have this problem here:
Given the series
[tex]\sum_{n=1} ^{\infty} \frac{x^{n+1}}{n(n+1)}[/tex]
show that this converges for every [tex]x \in \{ w \in \mathbb{C} \| \|w \| \leq 1 \}[/tex]
Solution:
Since
[tex] \sum_{n=1} ^{\infty} \left| \frac{x^{n+1}}{n(n+1)} \right| = \sum_{n=1} ^{\infty} \frac{1}{n(n+1)} \right| [/tex]
Since [tex]\sum_{n=1} ^{\infty} \frac{1}{n(n+1)} \right| [/tex] convergence then according the definition then
[tex]\sum_{n=1} ^{\infty} \frac{x^{n+1}}{n(n+1)}[/tex] is absolute convergent.
Thus
[tex]x \in \{ w \in \mathbb{C} \| \|w \| \leq 1 \}[/tex]. Right?
Best Regards
Fred
Last edited: