Prove that $\sigma_k$ is a multiplicative function For each real $k$,we define:
$$\sigma_k(n)=\sum_{d \mid n} d^k$$
$$\text{Prove that } \sigma_k \text{ is a multiplicative function.}$$
That's what I have tried:


*

*$$\sigma_k(1)=\sum_{d \mid 1} d^k=1$$

*Now,we have to show that if $(m,n)=1$,then we have $\sigma_k(m \cdot n)=\sigma_k(m) \sigma_k(n)$
At the case when one of $m,n$ is $1$,it is obvious.
Let $m,n>1$:
$$\text{We know that if } (m,n)=1 , d_1 \text{ goes through all the positive divisors of  } m \text{ and } d_2 \text{ goes through all the positive divisors of } n, \text{ then } d_1 \cdot d_2 \text{ goes throught the positive divisors of } mn.$$
So, $$\sigma_k(mn)=\sum_{d_1 \mid m , d_2 \mid n} (d_1 d_2)^k=\sum_{d_1 \mid m} d_1^k \cdot \sum_{d_2 \mid n} d_2^k=\sigma_k(m) \sigma_k(n)$$
Therefore,the function is multiplicative.
Could you tell me if it is right?
 A: Yes, this is correct. But I'd like to add a slightly different way of viewing this proof (and similar proofs).
Call the (Dirichlet) convolution of the arithmetic functions $f(n)$ and $g(n)$ to be
$$ f \star g (n) := \sum_{d \mid n} f(d)g\left(\frac nd\right),$$
which can also be written as a product across the various ways of writing $n = ab$:
$$ f \star g (n) = \sum_{ab = n} f(a)g(b).$$
Notice that $\displaystyle \sum_{d \mid n} f(d)$ is the convolution of $f(n)$ and $1(n)$, where by $1(n)$ I mean the function that sends everything to $1$. (This is how this ties back to your question).
Claim: if both $f$ and $g$ are multiplicative functions, then $f \star g (n)$ is also multiplicative. 
Proof: Proceed directly, just as you did in your original post. Write $n = AB$, where $\gcd(A,B) = 1$. Then 
$$\begin{align}
\displaystyle f\star g (AB) &= \sum_{d \mid n} f(d)g(n/d) \\
&= \sum_{\substack{n = AB \\ a \mid A \\ b \mid B}}f(ab)g(AB/ab) \\
&= \sum_{\substack{a\mid A\\ b\mid B}} f(a)f(b)g(A/a)g(B/b) \\
&= f\star g(A) \cdot f\star g(B).\end{align}$$
That's what we wanted to show. $\diamondsuit$
Now the raise-to-the-$k$th-power-function $(\cdot)^k: n \mapsto n^k$ is clearly totally multiplicative. The function $1(\cdot): n \mapsto 1$ is also totally multiplicative. Thus $1 \star (\cdot)^k (n)$, which is $\sigma_k(n)$, is multiplicative. 
And a vast array of multiplicative functions are realized as the Dirichlet convolution of multiplicative functions. Aside: this is a nice path into the study of analytic number theory and Dirichlet series, as Dirichlet convolution appears as the $n$th coefficients of products of Dirichlet series. A lot of interesting data is held in the coefficients of Dirichlet series.
