Prove that:
[tex]\sum_{x=1}^{\infty }\frac{1}{n^{x}} = \frac{1}{n-1}[/tex]
[tex]\sum_{x=1}^{\infty }\frac{1}{(-n)^{x}} = -\frac{1}{n+1}[/tex]
with all [tex]n[/tex] [tex](n \ne 0,n \in ℕ)[/tex]
Guest wrote:Prove that:
[tex]\sum_{x=1}^{\infty }\frac{1}{n^{x}} = \frac{1}{n-1}[/tex]
[tex]\sum_{x=1}^{\infty }\frac{1}{(-n)^{x}} = -\frac{1}{n+1}[/tex]
with all [tex]n[/tex] [tex](n \ne 0,n \in ℕ)[/tex]
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