$1-{1\over 2}+{1\over 3}-{1\over 4}+{1\over 5}-{1\over 6}+\dots-{1\over 2012}+{1\over 2013}$ The sum $1-{1\over 2}+{1\over 3}-{1\over 4}+{1\over 5}-{1\over 6}+\dots-{1\over 2012}+{1\over 2013}$ is equal,
a) ${1\over 1006}+{1\over 1007}+{1\over 1008}+\dots+{1\over 2013}$
b) ${1\over 1007}+{1\over 1008}+{1\over 1009}+\dots+{1\over 2013}$
c)${1\over 1006}+{1\over 1007}+{1\over 1008}+\dots +{1\over 2012}$
d) ${1\over 1007}+{1\over 1008}+{1\over 1009}+\dots+{1\over 2012}$
I tried by separating negatives and positives terms but did not get anything nice and simpler. Thank you for helping.

$$\sum_{1\le r\le 2n+1}(-1)^{r-1}\frac1r$$
$$=\sum_{1\le r\le 2n+1}\frac1r-2\left(\sum_{1\le r\le n-1}\frac1{2r}\right)$$
$$=\sum_{1\le r\le 2n+1}\frac1r-\left(\sum_{1\le r\le n-1}\frac1{r}\right)$$
$$=\sum_{n\le r\le 2n+1}\frac1r$$
Here $2n+1=2013\implies n=?$
