What does ( 0.25)^(3n) converge towards? (EASY Q)

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Homework Help Overview

The discussion revolves around the convergence of the series defined by the function f(x) = 0.25^(3n) for natural numbers n ranging from 0 to infinity. Participants are exploring the summation of this series and its properties.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the nature of the series, with one suggesting it resembles a geometric series. Questions arise regarding the appropriate formula for summation, particularly for different starting points of n.

Discussion Status

There is an active exploration of the series' characteristics, with some participants providing insights into geometric series and relevant summation formulas. However, no consensus has been reached on the specific summation value or the implications of the series' convergence.

Contextual Notes

Participants are navigating the rules of summation and the conditions under which the series converges, particularly focusing on the starting index of n and the behavior of the series as n approaches infinity.

polosportply
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Really quick question here:

I want to find the sommation of f(x) = 0.253n for x being a Natural ( N) number going from 0 to infinity.

k1k2n converges towards what, as a general rule?

Where k1, k2 are constants and n= 0,1,2,3 ...

EDIT: NO, not what it converges to, but what the sommation is equal to... as in f(0)+f(1) +f(2) + ...
 
Last edited:
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It's a geometric series isn't it? It's ((1/4)^3)^n. What do you know about the sum of geometric series?
 
Well, I'm looking for = ∑rn-1 for n going from 1 to infinity... But isn't there a formula for where n starts at 0?

I also know that you can play around sith Sn= (1-rn) / (1-r) , but I don't remember/know where this leads to.
 
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I'll help you out. The sum for n=0 to infinity of r^n is 1/(1-r) for r<1. That's the n->infinity limit of your Sn.
 
All right, thanks a lot
 
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