Show that $\sum_{n>Y} \frac{\tau(n) }{n^2} =\frac{\log Y+C+2}{Y} +O(1/Y^{3/2})$

Question

The following question is from my number theory assignment and I am unable to completely solve the problem. I have been following Dekonick and Luca.

Let $\tau(n)=\sum_{d\mid n} 1$ be the divisor function. Using $\sum_{n\leq x} \tau(n) =x(\log x+C) +O(x^{1/2})$, show that $\sum_{n>Y} \frac{\tau(n) }{n^2} =\frac{ \log Y+C+2}{Y} +O(1/Y^{3/2})$.

Attempt: I wrote $\sum_{n>Y} \frac{\tau(n)}{n^2}= -\sum_{n\leq Y}\frac{\tau(n)}{n^2} + \sum_{n\geq 1} \frac{\tau(n) }{n^2}$.

I calculated $\sum_{n\leq Y}\frac{\tau(n)}{n^2}= 2C+2 -\frac{2C+1 +\log Y }{Y}+ O(\frac{1}{Y^{3/2}})$. Hope I am right!

But I am unable to think which result to use to compute $\sum_{n\geq 1} \frac{\tau(n) }{n^2}$.

Can you please help with that?

Answer 1

Let $S(x)=\sum_{n\leq x} \tau(n) =x(\log x+C) +O(x^{1/2})$ and let $M>Y$. Summation by parts gives:

$$\sum_{n=Y}^{M} \frac{\tau(n) }{n^2}=\frac{S(M)}{M^2}-\frac{S(Y)}{Y^2}+\int_Y^M\frac{2S(x)dx}{x^3}.$$

But using the formula for $S(x)$ we have that

$$\int_Y^M\frac{2S(x)dx}{x^3}=\int_Y^M\frac{2(\log x+C)dx}{x^2}+O(Y^{-3/2}).$$

and integrating by parts the first term we get

$$\int_Y^M\frac{2(\log x+C)dx}{x^2}=\frac{2\log Y+2C+2}{Y}+O\!\left(\frac{\log M}{M}\right),$$

while $$\frac{S(M)}{M^2}=O\!\left(\frac{\log M}{M}\right),\qquad -\frac{S(Y)}{Y^2}=-\frac{\log Y+C}{Y}+O(Y^{-3/2}).$$

Putting the above together and letting $M \to +\infty$ the result follows.

Answer 2

Ok, since you asked in the comments but it's a bit long for there:

To compute $\sum _{n\geq 1}\tau (n)/n^s$. If you have two arithmetic functions $f,g$ then their Dirichlet convolution is defined by $$ (f\star g)(n)=\sum _{dd'=n}f(d)g(d').$$ The Dirichlet series of $f\star g$ is then given by \begin{align*}\text {series of $f\star g$} & =\sum _{n\geq 1}\frac {(f\star g)(n)}{n^s}=\sum _{n,d,d'=1\atop {dd'=n}}\frac {f(d)g(d')}{n^s}=\sum _{d,d'=1}^\infty \frac {f(d)g(d')}{(dd')^s}\\ &=\left (\sum _{d=1}^\infty \frac {f(d)}{d^s}\right )\left (\sum _{d'=1}^\infty \frac {g(d')}{d'^s}\right )=\text { series of $f$} \times \text { series of $g$}.\end{align*} In other words, the Dirichlet series of the convolution is the product of the individual Dirichlet series. I'm ignoring convergence issues because I think first maybe it's good just to make sure that idea is there (and because I am lazy).

This is the concept you need to have understood for your question I think.