Prove without induction : $\sum_{k=1}^{2n} \frac{(-1)^{k+1}}{k} = \sum_{k=1}^n \frac{1}{k+n}$

Question

Prove without induction that : $$ \sum_{k=1}^{2n} \frac{(-1)^{k+1}}{k} = \sum_{k=1}^n \frac{1}{k+n} $$

Please if you have any elementary tricks just post hints.

Have you checked to see whether this is even true for, say, $n=1$? — Gerry Myerson, Feb 18 '13 at 12:03

score 5 · Accepted Answer · edited Apr 13 '17 at 12:20

5

Rewriting $\displaystyle H(2n)-H(n)=\sum_{k=1}^n\frac1{k+n}$, the first part of this answer becomes: $$ \begin{align} \sum_{k=1}^n\frac1{k+n} &=\sum_{k=1}^{2n}\frac1k-\sum_{k=1}^n\frac1k\\ &=\sum_{k=1}^n\frac1{2k-1}+\frac1{2k}-\frac1k\\ &=\sum_{k=1}^n\frac1{2k-1}-\frac1{2k}\\ &=\sum_{k=1}^{2n}(-1)^{k-1}\frac1k \end{align} $$ which is the desired equation.

edited Apr 13 '17 at 12:20

Community

1

answered Feb 18 '13 at 13:49

robjohn

345,667

4

Nice (+1). This is actually Botez-Catalan identity. – user 1591719 Feb 18 '13 at 18:02
Great! Another name to remember... ;-) – robjohn Feb 18 '13 at 18:06

score 2 · Answer 2 · answered Feb 18 '13 at 12:58

2

$$ \sum_{k=1}^{2 n} \frac{(-1)^{k+1}}{k} = \sum_{k=1}^{n} \frac{(-1)^{k+1}}{k} +\sum_{k=1}^{n} \frac{(-1)^{k+n+1}}{k+n} = \sum_{k=1}^{n}(-1)^{k+1} \left(\frac{1}{k}+ \frac{(-1)^{n}}{k+n}\right) = \sum_{k=1}^{n}(-1)^{k+1} \frac{(-1)^{n}k+k+n}{k(k+n)} $$ and then consider two cases: $n$ is odd and $n$ is even.

answered Feb 18 '13 at 12:58

Boris Novikov

17,470

assuming its odd how should we deal with : $\sum \frac{(-1)^{k+1}n}{k(k+n)} $ ? – Tulip Feb 18 '13 at 13:10
$\frac{n}{k(k+n)}=\frac{1}{k}-\frac{1}{k+n}$ – Boris Novikov Feb 18 '13 at 13:46

score 2 · Answer 3 · answered Feb 18 '13 at 13:31

The left hand side has twice as much terms as the right hand side, so you can combine some terms of the left hand to match a term on the right. If you write the right hand side as $\sum_{k=n+1}^{2n}\frac1k$, then some terms (those with odd $k$) also occur on the left so you can just leave them; the others (with even $k$) have the opposite sign as the ones on the left, so they need to match a combination of terms. For instance for $n=4$ you have $$ \frac11-\frac12+\frac13-\frac14+\frac15-\frac16+\frac17-\frac18 =\frac15+\frac16+\frac17+\frac18, $$ the $\frac15$ and $\frac17$on both sides match, the term $\frac16$ on the right mathes the combination $\frac13-\frac16$ on the left, and the final term $\frac18$ on the right matches the combination $\frac11-\frac12-\frac14-\frac18$. Can you spot the pattern?

score 0 · Answer 4 · answered Feb 18 '13 at 17:43

Hint $\rm\ \ a_k = \dfrac{1}k\, =\, 2\left(\dfrac{1}{2k}\right)\, =\, 2\,a_{2k}\:$ $\Rightarrow$ $\rm\:a_{2k}-a_{k}= -a_{2k},\:$ which, in the bottom 2 rows below yields

$\rm\ \ a_n\! + \cdots + a_{2n} = \ \begin{array}{l} \rm \ \ \ a_1 + a_3 + \cdots + a_{2n-1}\\ \rm \!+\, a_2 + a_4 + \cdots + a_{2n}\\ \rm -a_1 - a_2 + \cdots -a_n \end{array} \!\!=\ \begin{array}{l} \rm \ \ \ a_1 + a_3 + \cdots + a_{2n-1}\\ \rm -a_2 - a_4 + \cdots -a_{2n} \\ \rm\ \ since\ \ a_k\! = 2 a_{2k} \end{array} \!=\ a_1\! - a_2 + a_3 -\cdots - a_{2n} $

Prove without induction : $\sum_{k=1}^{2n} \frac{(-1)^{k+1}}{k} = \sum_{k=1}^n \frac{1}{k+n}$

4 Answers4