Summation bounds for the following problem

Question

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 ${2^{n} -1}$.

Answer 1

Changing the bottom index simply omits the first term in the sum, which is $\binom{n}{0}=1$. So $$\displaystyle\sum_{k=1}^n\binom nk=\displaystyle\sum_{k=0}^n\binom nk-1.$$

Answer 2

Isn’t it just that $$\sum_{k=0}^n c_k=c_0 +\sum_{k=1}^n c_k$$ and noting that your $c_0=1$?