Sum of convergent series

steve.b

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Sep 26, 2010
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How to prove that n=1N5n=N4\displaystyle \sum_{n=1}^{\infty}\frac{N}{5^n}=\frac{N}{4}?
What about: n=1N5n=?\displaystyle \sum_{n=1}^{\infty}\left\lfloor \frac{N}{5^n} \right\rfloor=?
I think it's the same that n=1log5NN5n=?\displaystyle \sum_{n=1}^{\log_5 N}\left\lfloor \frac{N}{5^n} \right\rfloor=?

Can anybody help me?
Thank you in advanve!
 
steve.b said:
How to prove that n=1N5n=N4\displaystyle \sum_{n=1}^{\infty}\frac{N}{5^n}=\frac{N}{4}?
What about: n=1N5n=?\displaystyle \sum_{n=1}^{\infty}\left\lfloor \frac{N}{5^n} \right\rfloor=?
I think it's the same that n=1log5NN5n=?\displaystyle \sum_{n=1}^{\log_5 N}\left\lfloor \frac{N}{5^n} \right\rfloor=?

Can anybody help me?
Thank you in advanve!

(1/5)[sup:36ojifp4]n[/sup:36ojifp4] is a geometric series wher S[sub:36ojifp4]n[/sub:36ojifp4] = [1-(1/5)[sup:36ojifp4]n[/sup:36ojifp4]]/[1-1/5]

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