# Summation bounds for the following problem

Evaluate $$\displaystyle\sum_{k=1}^n \left(\frac{{n-1 \choose k-1}}{k}\right)$$

Now the way I solved this was by multiplying by n and turning it into

$$\displaystyle\sum_{k=1}^n\binom nk$$

my mistake was that I took the summation when $$k=0$$ to give me an easy $$2^{n}$$ which I divide by n to get a final answer of

$${2^{n} \over n}$$ when it is actually supposed to be $${2^{n} -1 \over n}$$

My question is how would I find that changing $$\displaystyle\sum_{k=1}^n\binom nk$$ to $$\displaystyle\sum_{k=0}^n\binom nk$$ would change the numerator to $${2^{n} -1}$$.

• You seem to be missing a factor $\tfrac1n$ in your summations. – Servaes Mar 15 at 22:48

Changing the bottom index simply omits the first term in the sum, which is $$\binom{n}{0}=1$$. So $$\displaystyle\sum_{k=1}^n\binom nk=\displaystyle\sum_{k=0}^n\binom nk-1.$$
• alright so since it omits the first term which is equal to 1, i would just subtract 1 from the summation. That make sense. Now sorry if this is a really obvious question, but I am completely new to this, $\binom{n}{0}=1$ is this from pascals triangle? or is just whenever you "choose" 0 its always equal to 1? – Brownie Mar 15 at 22:49
• There are many ways to see it, and what is the easiest way depends on your definition. You could read it off from Pascals triangle; these are all the $1$'s at the beginning of every row. Or from the combinatiorial interpretation; there is precisely $1$ way to choose nothing from $n$ elements. Or algebraically $$\binom n0=\frac{n!}{0!n!}=1.$$ – Servaes Mar 15 at 22:51
Isn’t it just that $$\sum_{k=0}^n c_k=c_0 +\sum_{k=1}^n c_k$$ and noting that your $$c_0=1$$?