Let p be a prime number. Decide $$\sum_{k=0}^n \phi(p^k)$$
I have had difficulties with how to deal with the problem. Any help is much appreciated.
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3Just exploit $\phi(p^k)=p^k-p^{k-1}$. Your sum is a telescopic one. – Jack D'Aurizio Sep 29 '19 at 16:28
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Bearing in mind that $\phi(p^0)=\phi(1)=1$, not $1-\frac{1}{p}$ – Keith Backman Sep 30 '19 at 15:02