I am looking for symptotic formula for $\sum_{n<x} \frac {d(n)}{\sqrt{n}}$ which doesn't use $\zeta(\frac{1}{2})$
My guess is – perhaps it is something like $A\sqrt{x}\log{x} + B \sqrt{x} + O(\log{x})$ where $A, B$ are some constants, based on these formulas:
$$\sum_{n<x} \frac {d(n)}{n} = \frac{1}{2}(\log{x})^2 + 2\gamma\log{x} + O(1)$$
$$\sum_{n<x} {d(n)} = x\log{x} + (2\gamma-1)x + O(\sqrt{x})$$
Per Huxley the error term can be improved to
$$\sum_{n<x} {d(n)} = x\log{x} + (2\gamma-1)x + O(x^\theta)$$ with $ \inf \theta \le 131/416 = 0.31490384615 $
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Best Answer
Use Abel's summation formula.
If $$A(x) = \sum_{n< x} d(n) = x \log x +(2\gamma -1) x + O(\sqrt{x})$$
then $$\sum_{n< x}\frac{d(n)}{\sqrt{n}} = \frac{A(x)}{\sqrt{x}} + \int_1^x\frac{A(t)}{2\ t^{3/2}}\ dt$$
From this you should be able to get
$$\sum_{n< x}\frac{d(n)}{\sqrt{n}} = 2\sqrt{x}\log x + 4(\gamma -1)\sqrt{x}+O(\log x)$$