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This may be a basic calculus question but I just can not understand a point in the following argument:

We have the following equality($d$ denotes the divisor function): $\displaystyle\sum_{k \leq x}d(k)=xlogx+(2C-1)x+O(\sqrt{x})$ where $C=\displaystyle \lim_{n \to \infty}(1+\cfrac{1}{2}+\cfrac{1}{3}+\dots+\cfrac{1}{n}-logn)$.

Now, we say that $f(x)$ is asymptotic to $g(x)$ if $\displaystyle\lim_{x \to \infty}\cfrac{f(x)}{g(x)}=1$ and write $f(x)$~$g(x)$.

So, Apostol says that $\displaystyle\sum_{k \leq x}d(k)$~$xlogx$.

However, can we also say that $\displaystyle\sum_{k \leq x}d(k)$~$(2C-1)x$ for example?

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    No, $x\log x$ grows faster than $(2C-1)x$. – 2017-01-14
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    I can see that it grows faster but we are taking the limit as $x$ goes to $\infty$? – 2017-01-14
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    Yes, and "grows faster" means $\lim\limits_{x\to\infty} \frac{x\log x}{(2C-1)x} = +\infty$. We have $$\frac{\sum\limits_{k\leqslant x} d(k)}{(2C-1)x} = \frac{\log x}{2C-1} + 1 + O(1/\sqrt{x}),$$ and the first term tends to $+\infty$. – 2017-01-14
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    Okay, I think I am missing some point, let me work on it. Thank you. – 2017-01-14

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