This is the homework, and it shouldn't be difficult, but I can't find the proper identity that would help me simplify this sum:
$$\sum_{n=0}^m \frac{1}{n+1}\binom{m}{n}$$
Through calculating the results, I can see that the simplified version is:
$$\frac{2^m-1}{m+1}$$
But I don't know how to transform the former into the later. You need not give the complete solution (although, that's welcomed too), but the identities needed for the simplification should suffice.
EDIT:
How I counted:
$\frac{m!}{(n+1)!(m-n)!}$ repeated $m$ times while $n$ increases from 0 to $m$. You can also see the code here: http://pastebin.com/RJ9jd966
Best Answer
Hint. Consider the function $$ \sum_{n=0}^m\frac{1}{n+1}{m\choose n} x^{n+1}.\tag{1} $$ Differentiating it gives you something very familiar.
Since you've already accepted an answer, I'll work out my answer so that you can see another approach. Differentiating the above polynomial gives $$ \sum_{n=0}^m{m\choose n}x^n $$ which may be recognized as a special case of Newton's binomial series: it is equal to $(x+1)^m$. The primitive of this polynomial is $p(x)=\frac{1}{m+1}(x+1)^{m+1}+c$, where we have to choose the constant such that $p$ agrees with (1). One way to do this, is by looking at the value at $0$. $$ p(0)=\frac{1}{m+1}+c $$ while the function in (1) gives $0$ at $0$. This means that we must choose $c=-\frac{1}{m+1}$. So we obtain $$ \sum_{n=0}^m\frac{1}{n+1}{m\choose n} x^{n+1}=\frac{(x+1)^{m+1}}{m+1}-\frac{1}{m+1} $$ evaluating the latter term at $1$ gives the answer: $$ \frac{2^{m+1}-1}{m+1}. $$