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\centerline
{
\bf Solutions to the Attendance Quiz \# 17 for Dr. Z.'s Number Theory
Course for Nov. 4, 2013
}
\bigskip

{\bf 1.} State the formula for  $\sigma_1(n)$, and verify it for $n=20$.

{\bf Sol. to 1}: If
$$
n= \prod_{i=1}^{k} p_i^{\alpha_i} \quad,
$$
Then $\sigma_1(n)$ (aka as $\sigma(n)$) is computed by
$$
\sigma(n)= \prod_{i=1}^{k} (1+ p_i + (p_i)^2 + \dots + (p_i)^{\alpha_i}) \quad .
$$
For $n=20$ we have
$$
n= 2^2 \cdot 5^1 \quad ,
$$
so
$$
\sigma(20)= (1+2+4) \cdot (1+5)= 7 \cdot  6 = 42 \quad .
$$
Now let's do it {\bf from the definition}.
$$
Div(20)=\{ 1,2,4 \} \cdot \{ 1, 5 \}=
\{ 1, 2, 4, 5, 10, 20 \}
$$
Adding the members of $Div(20)$ together gives:
$$
\sigma(20)= 1+2+4+5+10+20= 42 \quad .
$$
Yea! The formula agrees with the definition, at least for $n=20$.

{\bf 2.} Verify the Dirichlet series for $\sigma(n)$ for up to $n=4$

{\bf Sol. to 2.}
$$
\zeta(s) \zeta(s-1)=
(1+ \frac{1}{2^s} + \frac{1}{3^s} + \frac{1}{4^s} + \dots )
(1+ \frac{2}{2^s} + \frac{3}{3^s} + \frac{4}{4^s} + \dots )
$$
$$
1+ \frac{2}{2^s} + \frac{3}{3^s} + \frac{4}{4^s} + \dots 
$$
$$
+\frac{1}{2^s} + \frac{2}{4^s} + \dots
$$
$$
+\frac{1}{3^s} + \dots
$$
$$
+\frac{1}{4^s} + \dots
$$
(Note, since we are only worried about $\sigma(n)$ up to $n=4$, every
time we see $1/5^s, 1/6^s, \dots$ etc. we replace it by $\dots$).

So this equals (collecting terms)
$$
1+ \frac{3}{2^s} + \frac{4}{3^s} + \frac{7}{4^s} + \dots 
$$
and indeed $\sigma(1)=1$, $\sigma(2)=1+2=3$, $\sigma(3)=1+3=4$, $\sigma(4)=1+2+4$.

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