Exercise 4.3.12 ($\frac{S(n)}{n^2} = \sum_{d \mid n} \frac{1}{6} \mu(d) \left( \frac{2n}{d} + 3 + \frac{d}{n} \right).$)

Question

Combine the results of the two preceding problems to get
$$\sum_{d \mid n} \frac{S(d)}{d^2} = \frac{1}{6} \left( 2n + 3 + \frac{1}{n} \right).$$
Then apply the Möbius inversion formula to get
$$\frac{S(n)}{n^2} = \sum_{d \mid n} \frac{1}{6} \mu(d) \left( \frac{2n}{d} + 3 + \frac{d}{n} \right).$$

Richard Ganaye · Accepted Answer

\begin{proof} 
Combining the results of the two preceding problems, we get
$$n^2 \sum_{d \mid n} \frac{S(d)}{d^2} =\sum_{j=1}^n j^2 =  \frac{n(n+1)(2n+1)}{6}.$$
Therefore
\begin{align*}
 \sum_{d \mid n} \frac{S(d)}{d^2} &= \frac{(n+1)(2n+1)}{6n}\
 &=\frac{1}{6} \left( 2n + 3 + \frac{1}{n} ight).
 \end{align*}
 Put $f(n) = \frac{S(n)}{n^2}$ for every positive integer $n$, and $F(n) = \sum_{d \mid n} f(d)$. Then  $F(n) = \frac{1}{6} \left( 2n + 3 + \frac{1}{n} ight)$, and the Möbius inversion formula gives
\begin{align*}
\frac{S(n)}{n^2}&=f(n)  \
&= \sum_{d \mid n} \mu(d) F\left( \frac{n}{d} ight)\
&= \sum_{d \mid n} \mu(d) \frac{1}{6}  \left( \frac{2n}{d} + 3 + \frac{d}{n} ight).
\end{align*}
So
$$\frac{S(n)}{n^2} = \sum_{d \mid n} \frac{1}{6} \mu(d) \left( \frac{2n}{d} + 3 + \frac{d}{n} ight).$$
\end{proof}

Exercise 4.3.12 ($\frac{S(n)}{n^2} = \sum_{d \mid n} \frac{1}{6} \mu(d) \left( \frac{2n}{d} + 3 + \frac{d}{n} \right).$)

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Exercise 4.3.12 ($\frac{S(n)}{n^2} = \sum_{d \mid n} \frac{1}{6} \mu(d) \left( \frac{2n}{d} + 3 + \frac{d}{n} \right).$)

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