Summation of a finite series Let
$$f(n) = \frac{1}{1} + \frac{1}{2} + \frac{1}{3}+ \frac{1}{4}+...+ \frac{1}{2^n-1} \forall \ n \ \epsilon \ \mathbb{N} $$
If it cannot be summed , are there any approximations to the series ?
 A: $$f(n)\approx \ln(2^n-1)+\gamma\approx n\ln 2+\gamma$$
A: $$1+(\frac12+\frac12)+ (\frac14+\frac14+\frac14+\frac14)+ \dots (\dots \frac1{2^{n-1}})\gt$$
$$\frac{1}{1} + \frac{1}{2} + \frac{1}{3}+ \frac{1}{4}+...+ \frac{1}{2^n-1}\gt$$  $$1+\frac12+(\frac14+\frac14)+(\frac18+\frac18+\frac18+\frac18)+\dots +( \dots \frac1{2^n})$$ (compare equivalent terms, and count the elements in the last bracket carefully) gives some bounds (the summation is straightforward). Since $\frac 1n$ is monotone decreasing, it is easy to use the integral to give tighter bounds than this.
The question is tagged as a contest, so integration may not be appropriate (depends on contest parameters). On the other hand the sums above, though rather crude, can be computed in a contest, and make sense of the relationship of the sum with a power of $2$. 
I'd be interested in a better elementary estimate not involving integration.
A: $f(n)=H_{2^n-1}$, the $(2^n-1)$-st harmonic number. There is no closed form, but link gives the excellent approximation
$$H_n\approx\ln n+\gamma+\frac1{2n}+\sum_{k\ge 1}\frac{B_{2k}}{2kn^{2k}}=\ln n+\gamma+\frac1{2n}-\frac1{12n^2}+\frac1{120n^4}-\ldots\;,$$
where $\gamma\approx 0.5772156649$ is the Euler–Mascheroni constant, and the $B_{2k}$ are Bernoulli numbers.
