Sum of convergent series

[steve.b]

How to prove that \(\displaystyle \sum_{n=1}^{\infty}\frac{N}{5^n}=\frac{N}{4}\)?
What about: \(\displaystyle \sum_{n=1}^{\infty}\left\lfloor \frac{N}{5^n} \right\rfloor=?\)
I think it's the same that \(\displaystyle \sum_{n=1}^{\log_5 N}\left\lfloor \frac{N}{5^n} \right\rfloor=?\)

Can anybody help me?
Thank you in advanve!

 

[Deleted member 4993]

How to prove that \(\displaystyle \sum_{n=1}^{\infty}\frac{N}{5^n}=\frac{N}{4}\)?
What about: \(\displaystyle \sum_{n=1}^{\infty}\left\lfloor \frac{N}{5^n} \right\rfloor=?\)
I think it's the same that \(\displaystyle \sum_{n=1}^{\log_5 N}\left\lfloor \frac{N}{5^n} \right\rfloor=?\)

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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[steve.b]

Ok, thank you for your help, I coped with it.