I came across this question:
If $\pi(n)$ denote the Euler function. What's the radius of convergence of this power series? $$ \sum\limits_{n=2}^{\infty} \pi(n) z^{n}$$
Any hint would be appreciated.
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6I never heard of $\pi(n)$ as "the Euler function". Is it the totient function or the prime-counting function? – Jack D'Aurizio Aug 22 '14 at 23:02
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@JackD'Aurizio Yes, is the prime-counting function. – felipeuni Aug 22 '14 at 23:04
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1Either way, the radius of convergence is $1$. – Thomas Andrews Aug 22 '14 at 23:05
2 Answers
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The radius of convergence is $1$ without knowing anything deep at all about prime counting. For $n\ge2$, we have $1\le\pi(n)\le n$. This gives the bounds, for $x\ge0$, $$ \frac x{1-x}=\sum_{n=1}^\infty1x^n\le x+\sum_{n=2}^\infty\pi(n)x^n\le\sum_{n=1}^\infty nx^n=\frac{x}{(1-x)^2} $$ and the series on both sides have radius of convergence $1$.
robjohn
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The radius of convergence is one since by the Chebyshev theorem (the weak version of the PNT) $$\frac{n}{\log n}\ll \pi(n)\ll\frac{n}{\log n}.$$
Jack D'Aurizio
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- Should the RHS and LHS of that inequality be identical? 2) can we say anything about convergence on the unit circle?
– Semiclassical Aug 22 '14 at 23:43 -
@Semiclassical: Yes, Vinogradov's notation here just means that for two suitable constants $c_1,c_2$, for any $n$ big enough we have $c_1\frac{n}{\log n}\leq \pi(n)\leq c_2\frac{n}{\log n}$. – Jack D'Aurizio Aug 22 '14 at 23:45
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We do not have convergence on the unit circle since the general term $\frac{n}{\log n}, z^n$ is not converging to zero when $n\to +\infty$. – Jack D'Aurizio Aug 22 '14 at 23:50