Infinite Series

Arithmetic and Geometric progressions.

Infinite Series

Postby Guest » Thu Nov 19, 2020 8:02 am

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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Re: Infinite Series

Postby HallsofIvy » Fri Jan 01, 2021 3:34 pm

Guest wrote:Prove that:
[tex]\sum_{x=1}^{\infty }\frac{1}{n^{x}} = \frac{1}{n-1}[/tex]

That isn't necessarily true! it requires that |n|< 1 and usually (not always) "n" represents a positive integer.
Do you know anything about "geometric series"?

A geometric series is one of the form [tex]\sum a^n[/tex]. It can be shown that for |a|< 1 that series converges. Suppose it converges to "S". That is, S= 1+ a+ a^2+ .... + a^n+ ... Then S- 1= a+ a^2+ ...+ a^n+ ...= a(1+ a+ ...+ a^(n-1)+ ...). But that last sum, since it goes to infinity, is again S. S- 1= aS so S- aS= (1- a)S= 1 and then S= 1/(1- a).

[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]

Let a, above, be -n instead of n.

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