how can it be proved that
1/1+1/2+1/3+... +1/n<k*(log of n to the base e),
where k is a sufficiently large positive costant and n is a positive integer.

Take note that log_{e}(n)=ln(n)

This appears to be related to the gamma constant.

\lim_{n\to {\infty}}\left(\sum_{k=1}^{n}\frac{1}{k}-ln(n)\right)

=\lim_{n\to {\infty}}\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3} +....+\frac{1}{n}-ln(n)\right)={\gamma}\approx {.577215664....}

This can be thought of the difference between the sum and the integral of 1/x.

=\lim_{n\to {\infty}}\left(\sum_{k=1}^{n}-\int_{1}^{n}\frac{1}{x}dx\right)

If we let {\gamma} be of secondary importance, we can write it as:

\sum_{k=1}^{n}\frac{1}{k}-\int_{1}^{n}\frac{1}{x}dx\approx{\gamma}

{\therefore}, \;\ \sum_{k=1}^{n}\frac{1}{k}\approx \int_{1}^{n}\frac{1}{x}dx+{\gamma}

If you Google the 'gamma constant', you may find your proof in detail if the hint I gave is not enough.

Here's another way:

Given that ln(a) is defined as the area under the curve y=1/x from x=1 to x=a, it's straight forward to show that:

ln(N) > 1/2 + 1/3 +1/4 +... +1/N (see figure)

Or:

1+ln(N) > 1+ 1/2 + 1/3 +... +1/N

Since 1+ ln(N) = [1+ln(N)]* ln(N)/ln(N)

If we let k = [1+ln(N)]/ln(N)

You have:

1+ln(N) = k*ln(N) > 1 + 1/2 +1/3 +1... +1/N