How to find the constant B in here?

[Math100]

Homework Statement

Prove that, for $x\geq 2$, $\sum_{n\leq x}\frac{d(n)}{n^2}=B-\frac{\log x}{x}+O(\frac{1}{x})$, where $B$ is a constant that you should determine.

Relevant Equations

If $x\geq 2$ and $\alpha>0, \alpha\neq 1$, then $\sum_{n\leq x}\frac{d(n)}{n^{\alpha}}=\frac{x^{1-\alpha}\log {x}}{1-\alpha}+\zeta (\alpha)^2+O(x^{1-\alpha})$.

Proof:

Let $x\geq 2$.
Observe that
\begin{align*}
&\sum_{n\leq x}\frac{d(n)}{n^2}=\sum_{d\leq x}\frac{1}{d^2}\sum_{q\leq \frac{x}{d}}\frac{1}{q^2}\\
&=\sum_{d\leq x}\frac{1}{d^2}(\frac{(x/d)^{1-2}}{1-2}+\zeta {2}+O(\frac{1}{(x/d)^{2}}))\\
&=\frac{x^{1-2}}{1-2}\sum_{d\leq x}\frac{1}{d}+\zeta (2)\sum_{d\leq x}\frac{1}{d^2}+O(x^{1-2})\\
&=\frac{x^{1-2}}{1-2}(\log {x}+C+O(\frac{1}{x}))+\zeta {2}(\frac{x^{1-2}}{1-2}+\zeta {2}+O(x^{-2}))+O(x^{1-2}))\\
&=\frac{x^{1-2}\log {x}}{1-2}+C\cdot \frac{x^{1-2}}{1-2}+O(x^{-2})+\zeta (2)\cdot \frac{x^{1-2}}{1-2}+\zeta (2)^2+O(x^{-2})+O(x^{1-2})\\
&=\frac{x^{1-2}\log {x}}{1-2}+\zeta (2)^2+O(x^{1-2}).\\
\end{align*}
Thus $\sum_{n\leq x}\frac{d(n)}{n^2}=B-\frac{\log {x}}{x}+O(\frac{1}{x})$, where $B=\zeta (2)^2$.
Therefore, $\sum_{n\leq x}\frac{d(n)}{n^2}=\zeta (2)^2-\frac{\log {x}}{x}+O(\frac{1}{x})$ for $x\geq 2$.

 

[pasmith]

Is B = \zeta(2)^2 not sufficient, or do you need to use the standard result <br /> \zeta(2) = \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}?

 

[Math100]

Is B = \zeta(2)^2 not sufficient, or do you need to use the standard result <br /> \zeta(2) = \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}?

This was my question. How should I determine the constant $B$ in here? Is $\zeta (2)^2$ really sufficient? Or do I need to simplify/evaluate this value more, into $\frac{\pi^{2}}{6}$?

 

[fresh_42]

This was my question. How should I determine the constant $B$ in here? Is $\zeta (2)^2$ really sufficient? Or do I need to simplify/evaluate this value more, into $\frac{\pi^{2}}{6}$?

I would normally substitute it with $\pi^2/6$ in the final answer. But here we have a $B$ that swallows all constants wherever they come from, so there is no need for the substitution, a constant is a constant, no matter which. However, you should write $-1$ and not $1-2.$

 

[Math100]

I would normally substitute it with $\pi^2/6$ in the final answer. But here we have a $B$ that swallows all constants wherever they come from, so there is no need for the substitution, a constant is a constant, no matter which. However, you should write $-1$ and not $1-2.$

So $\zeta (2)^2=\zeta (4)=\sum_{n=1}^{\infty}\frac{1}{n^4}=\frac{\pi^{4}}{90}$? And $B=\frac{\pi^{4}}{90}$?

 

[fresh_42]

So $\zeta (2)^2=\zeta (4)=\sum_{n=1}^{\infty}\frac{1}{n^4}=\frac{\pi^{4}}{90}$? And $B=\frac{\pi^{4}}{90}$?

$\zeta (2)^2=\dfrac{\pi^4}{36}\neq \dfrac{\pi^4}{90}=\zeta (4).$

 

[Math100]

$\zeta (2)^2=\dfrac{\pi^4}{36}\neq \dfrac{\pi^4}{90}=\zeta (4).$

Why $\frac{\pi^{4}}{36}$?

 

[fresh_42]

Why $\frac{\pi^{4}}{36}$?

Because $\zeta(2)^2=\left(\dfrac{\pi^2}{6}\right)^2=\dfrac{\pi^4}{6\cdot 6}.$