Proving an Infinite Series Equation

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Gregg
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1. Problem



Prove that
[tex]\sum_{n=0}^{\infty} \frac{a}{k^n} = a\frac{k}{k-1}[/tex]



Homework Equations


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The Attempt at a Solution



Don't know how to do it at all.
 
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The infinite sum

[tex]\sum_{n=0}^\infty ar^n = \frac{a}{1-r}[/tex]

where |r|<1 (a necessary condition for convergence) is called the geometric series. (Putting r=1/k is your version.) One way to prove convergence and derive the value of the sum is to first derive the finite sum

[tex]\sum_{n=0}^N ar^n = a\frac{1-r^{N+1}}{1-r}[/tex]

and then take the limit.

You can find a simple proof of this latter sum at Wikipedia: http://en.wikipedia.org/wiki/Geometric_progression#Geometric_series

and (although redundant after having derived the finite sum) proof of the infinite sum there as well: http://en.wikipedia.org/wiki/Geometric_series#Sum .

I suggest scrolling slowly if you wish to exercise your mind and have a crack at it yourself.
 
Unco said:
The infinite sum

[tex]\sum_{n=0}^\infty ar^n = \frac{a}{1-r}[/tex]

where |r|<1 (a necessary condition for convergence) is called the geometric series. (Putting r=1/k is your version.) One way to prove convergence and derive the value of the sum is to first derive the finite sum

[tex]\sum_{n=0}^N ar^n = a\frac{1-r^{N+1}}{1-r}[/tex]

and then take the limit.

You can find a simple proof of this latter sum at Wikipedia: http://en.wikipedia.org/wiki/Geometric_progression#Geometric_series

and (although redundant after having derived the finite sum) proof of the infinite sum there as well: http://en.wikipedia.org/wiki/Geometric_series#Sum .

I suggest scrolling slowly if you wish to exercise your mind and have a crack at it yourself.


Thanks, I will. 7:30am though should sleep soon.