# How do I solve this summation [duplicate]

$$\sum_{b=1}^{n} \lfloor\frac{n}{b} \rfloor$$

Thanks!

Edit: I often come across summations I'm unable to solve. Is there some resource from which I can learn/practice solving complex summations. I hate asking numerous small questions of the same type repeatedly so if you could give some advice, I would be really grateful.

-

## marked as duplicate by user91500, Henning Makholm, Mark Bennet, Claude Leibovici, deinstApr 12 at 13:43

Does $\sum_{b=1}^n d(b)$ help, where $d(b)$ is the number of divisors of $b$? –  Hagen von Eitzen Apr 12 at 11:51
But the two summations are different aren't they. I don't think that will help –  Rushil Apr 12 at 11:53

It is likely that it doesn't have a nicer closed form. We can instead put it in some other meaningful form. Notice that $[\frac{n}{b}]$ is the number of multiples of $b$ that are less than or equal to $n$. So, in that sum, a number $k\leq n$ is going to be counted once for each one of its divisors. Let $\sigma(k)$ denote the number of divisors of $k$ then your sum should be equal to

$\sum_{k=1}^n\sigma(k)$

........

The problem is summing is not an easy one. Some nice books could be the book called $A=B$ which is available for free, and Concrete Mathematics.

-