$1-{1\over 2}+{1\over 3}-{1\over 4}+{1\over 5}-{1\over 6}+\dots-{1\over 2012}+{1\over 2013}$

Question

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.

possible duplicate of Is $\lim\limits_{k\to\infty}\sum\limits_{n=k+1}^{2k}{\frac{1}{n}} = 0$? — Jyrki Lahtonen, Nov 28 '13 at 08:02
Vote against closing. That an answer to another question answers this one doesn't mean it is automatically duplicate (what if I wanted to give a different answer to this question?). — Lord_Farin, Nov 28 '13 at 09:26

score 3 · Accepted Answer · answered Nov 28 '13 at 07:57

$$\sum_{1\le r\le 2n+1}(-1)^{r-1}\frac1r$$

$$=\sum_{1\le r\le 2n+1}\frac1r-2\left(\sum_{1\le r\le n-1}\frac1{2r}\right)$$

$$=\sum_{1\le r\le 2n+1}\frac1r-\left(\sum_{1\le r\le n-1}\frac1{r}\right)$$

$$=\sum_{n\le r\le 2n+1}\frac1r$$

Here $2n+1=2013\implies n=?$

1 Answers1