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

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In summary, the conversation discusses finding the summation of a function with a natural number going from 0 to infinity, where the function has constants and involves a geometric series. The formula for the sum of a geometric series is mentioned, as well as the limit for n->infinity. The conversation ends with gratitude for the help provided.
  • #1
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) + ...
 
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  • #2
It's a geometric series isn't it? It's ((1/4)^3)^n. What do you know about the sum of geometric series?
 
  • #3
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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  • #4
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.
 
  • #5
All right, thanks a lot
 
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1. What is the meaning of (0.25)^(3n)?

The expression (0.25)^(3n) represents the value of 0.25 raised to the power of 3n.

2. Does (0.25)^(3n) converge to a specific value?

Yes, as n approaches infinity, the value of (0.25)^(3n) will converge towards 0. This means that the value of the expression will get closer and closer to 0, but will never actually reach it.

3. How can (0.25)^(3n) be used in scientific calculations?

(0.25)^(3n) can be used to represent exponential decay, where the value decreases by a factor of 0.25 every time n increases by 1. This can be useful in fields such as biology and chemistry.

4. What is the relationship between (0.25)^(3n) and (0.25)^n?

The value of (0.25)^(3n) is equal to the value of (0.25)^n raised to the power of 3. In other words, (0.25)^(3n) is the cube of (0.25)^n.

5. Can (0.25)^(3n) ever equal 0?

No, since any number raised to the power of 0 is equal to 1, the value of (0.25)^(3n) will never be 0. However, as mentioned before, it will get infinitely close to 0 as n approaches infinity.

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