How can I find a bound for $\frac{n}{1} +\cdots+ \frac{n}{n}$ given $n< 10^{18}$? 
Could you help me find a bound for 
  $$\frac{n}{1} +\cdots+ \frac{n}{n}
$$
   given $n< 10^{18}$?

If there isn't any mathematical formula, algorithms will help too :)
 A: $$ \sum_{i=1}^{n} \frac{n}{i}=n \sum_{i=1}^{n} \frac{1}{i}=n H_n  $$ If you have a look to this paper you will find sharp bounds for the harmonic numbers and, in particular, equation $(6)$
$$\frac{1}{2 n+\frac{1}{1-\gamma }-2} \leq H_n-\log(n)-\gamma \lt\frac{1}{2 n+\frac 13}$$ giving 
$$n \left(\frac{1}{2 n+\frac{1}{1-\gamma }-2}+\log (n)+\gamma \right)\leq nH_n \lt n \left(\frac{1}{2 n+\frac{1}{3}}+\log (n)+\gamma \right)$$
The difference of the bounds is $$\Delta=\frac{(4-7 \gamma ) n}{(2 \gamma  (n-1)-2 n+1) (6 n+1)}$$ and, for $n=10^{18}$, $\Delta\approx 8.0\times 10^{-21}$ (quite small !).
So, for  $n=10^{18}$, the summation is $\approx 4.202375\times 10^{19}$ .
If fact, setting $n=10^k$, you should notice that $\Delta_k\approx  8.0\times 10^{-(k+3)}$ for any value of $k$.
This gives you, I hope, not only very sharp bounds but also very accurate values.
Edit
In this more recent paper, they give 
$$-\frac{1}{12 n^2+\frac{2 (7-12 \gamma )}{2 \gamma -1}}\leq H_n-\log(n)-\frac 1{2n} <-\frac{1}{12 n^2+\frac{6}{5}}$$ which, for $nH_n$ makes the difference of the bounds to be $$\Delta=\frac{(33 \gamma -19) n}{3 \left(10 n^2+1\right) \left(-6 n^2+12 \gamma 
   \left(n^2-1\right)+7\right)}$$ and, for $n=10^{18}$, $\Delta\approx 1.7\times 10^{-57}$.
Setting again $n=10^k$, you should notice that $\Delta_k\approx  1.7\times 10^{-3(k+3)}$ for any value of $k$.
A: Using the asymtopic behaviour of the harmonic series, we have
\begin{eqnarray*}
n \sum_{i=1}^{n} \frac{1}{i}= n \left(\ln n +\gamma+ \frac{1}{2n} + o(\frac{1}{n^2})\right)
\end{eqnarray*}
Which gives a value of about $\color{red}{4.1 \times 10^{19}}$.
A: Following is a very simple calculation.
For all $n>1$, $$\dfrac{n}{1} +\cdots+ \dfrac{n}{n}=n\sum\limits_{i=1}^n\frac{1}{i}\le n\int_{1}^n\frac{1}{x}~dx=n\ln n.$$
Since $n<10^{18}$, we have that
$$\dfrac{n}{1} +\cdots+ \dfrac{n}{n}<10^{18}\ln (10^{18})\approx 4.1\times 10^{19}.$$
