Show the following equality involving $\tau$, the number of divisors of the input.

Question

I want to show the following holds:

$\displaystyle \sum_{n=1}^{\infty} \frac{\tau (n)}{n^s} = \left( \sum_{n=1}^{\infty} \frac{1}{n^s} \right) ^2$

I tried to use induction but am not sure if that is the right approach as I struggle to show the result for $n+1$ (I tried using the fact that $\tau$ is multiplicative.) I don't know how else to start and proceed with the problem. Any help will be appreciated. I will list the properties of $\tau (n) (\tau$ is the number of divisors of the input) that I have tried to use. [Another consideration in my attempt involved using the last two properties. That is, if $f$ is multiplicative and $n = p_1^{e_1} \cdots p_r^{e_r} $ then $f(n) = f(p_1^{e_1} \cdots p_r^{e_r}) = f(p_1^{e_1})\cdots f(p_r^{e_r})]$.

$\displaystyle \tau (n) = \sum_{d|n} 1$.

$\tau (p^e) = e+1$.

$\tau$ is multiplicative.

If $n = p_1^{e_1} \cdots p_r^{e_r}$ then $\displaystyle \tau(n) = \prod_{i=1}^r (e_i + 1)$.

If $f$ is multiplicative then $\displaystyle F(n) = \sum_{d|n} f(d)$.

Answer 1

Let's use the definition and go forward:$$\sum_{n=1}^{\infty} \frac{\tau (n)}{n^s} $$$$=\sum_{n=1}^{\infty} \frac{\sum_{d|n}1}{n^s} $$$$=\sum_{n=1}^{\infty} \frac{1}{n^s}\sum_{d|n}1$$$$=\sum_{n=1}^{\infty} \sum_{d|n}\frac{1}{n^s}$$$$=\sum_{n=1}^{\infty} \sum_{d_1,d_2\\d_1d_2=n}\frac{1}{(d_1d_2)^s}$$$$=\sum_{d_1,d_2}\frac{1}{(d_1d_2)^s}$$$$=\sum_{d_1}\frac{1}{d_1^s}\sum_{d_2}\frac{1}{d_2^s}$$$$= \left( \sum_{n=1}^{\infty} \frac{1}{n^s} \right) ^2$$

Answer 2

$$ \begin{align} &\phantom{\ =}\left(\frac1{1^s}+\frac1{2^s}+\frac1{3^s}+\frac1{4^s}+\frac1{5^s}+\frac1{6^s}\right)\left(\frac1{1^s}+\frac1{2^s}+\frac1{3^s}+\frac1{4^s}+\frac1{5^s}+\frac1{6^s}\right)\\ &=\left(\color{#CCC}{\color{#C00}{\frac1{1^s}}+\frac1{2^s}+\frac1{3^s}+\frac1{4^s}+\frac1{5^s}+\frac1{6^s}}\right) \left(\color{#CCC}{\frac1{1^s}+\frac1{2^s}+\frac1{3^s}+\frac1{4^s}+\frac1{5^s}\color{#C00}{+\frac1{6^s}}}\right)\\ &+\left(\color{#CCC}{\frac1{1^s}}\color{#C00}{+\frac1{2^s}}\color{#CCC}{+\frac1{3^s}+\frac1{4^s}+\frac1{5^s}+\frac1{6^s}}\right) \left(\color{#CCC}{\frac1{1^s}+\frac1{2^s}}\color{#C00}{+\frac1{3^s}}\color{#CCC}{+\frac1{4^s}+\frac1{5^s}+\frac1{6^s}}\right)\\ &+\left(\color{#CCC}{\frac1{1^s}+\frac1{2^s}}\color{#C00}{+\frac1{3^s}}\color{#CCC}{+\frac1{4^s}+\frac1{5^s}+\frac1{6^s}}\right) \left(\color{#CCC}{\frac1{1^s}}\color{#C00}{+\frac1{2^s}}\color{#CCC}{+\frac1{3^s}+\frac1{4^s}+\frac1{5^s}+\frac1{6^s}}\right)\\ &+\left(\color{#CCC}{\frac1{1^s}+\frac1{2^s}+\frac1{3^s}+\frac1{4^s}+\frac1{5^s}\color{#C00}{+\frac1{6^s}}}\right) \left(\color{#CCC}{\color{#C00}{\frac1{1^s}}+\frac1{2^s}+\frac1{3^s}+\frac1{4^s}+\frac1{5^s}+\frac1{6^s}}\right)\\ &+\left(\color{#CCC}{\color{#090}{\frac1{1^s}}+\frac1{2^s}+\frac1{3^s}+\frac1{4^s}+\frac1{5^s}+\frac1{6^s}}\right) \left(\color{#CCC}{\frac1{1^s}+\frac1{2^s}+\frac1{3^s}\color{#090}{+\frac1{4^s}}+\frac1{5^s}+\frac1{6^s}}\right)\\ &+\left(\color{#CCC}{\frac1{1^s}\color{#090}{+\frac1{2^s}}+\frac1{3^s}+\frac1{4^s}+\frac1{5^s}+\frac1{6^s}}\right) \left(\color{#CCC}{\frac1{1^s}\color{#090}{+\frac1{2^s}}+\frac1{3^s}+\frac1{4^s}+\frac1{5^s}+\frac1{6^s}}\right)\\ &+\left(\color{#CCC}{\frac1{1^s}+\frac1{2^s}+\frac1{3^s}\color{#090}{+\frac1{4^s}}+\frac1{5^s}+\frac1{6^s}}\right) \left(\color{#CCC}{\color{#090}{\frac1{1^s}}+\frac1{2^s}+\frac1{3^s}+\frac1{4^s}+\frac1{5^s}+\frac1{6^s}}\right)\\ &+\left(\color{#CCC}{\color{#00F}{\frac1{1^s}}+\frac1{2^s}+\frac1{3^s}+\frac1{4^s}+\frac1{5^s}+\frac1{6^s}}\right) \left(\color{#CCC}{\frac1{1^s}+\frac1{2^s}\color{#00F}{+\frac1{3^s}}+\frac1{4^s}+\frac1{5^s}+\frac1{6^s}}\right)\\ &+\left(\color{#CCC}{\frac1{1^s}+\frac1{2^s}\color{#00F}{+\frac1{3^s}}+\frac1{4^s}+\frac1{5^s}+\frac1{6^s}}\right) \left(\color{#CCC}{\color{#00F}{\frac1{1^s}}+\frac1{2^s}+\frac1{3^s}+\frac1{4^s}+\frac1{5^s}+\frac1{6^s}}\right)\\[3pt] &+\cdots \end{align} $$ There are $4$ terms contributing to the term $\color{#C00}{\frac1{6^s}}$ in the product because $\tau(6)=4$.
There are $3$ terms contributing to the term $\color{#090}{\frac1{4^s}}$ in the product because $\tau(4)=3$.
There are $2$ terms contributing to the term $\color{#00F}{\frac1{3^s}}$ in the product because $\tau(3)=2$.