Prove without induction : $\sum_{k=1}^{2n} \frac{(-1)^{k+1}}{k} = \sum_{k=1}^n \frac{1}{k+n}$ 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.
 A: 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.
A: $$
\sum_{k=1}^{2 n} \frac{(-1)^{k+1}}{k} = \sum_{k=1}^{n} \frac{(-1)^{k+1}}{k} +\sum_{k=1}^{n} \frac{(-1)^{k+n+1}}{k+n} = \sum_{k=1}^{n}(-1)^{k+1} \left(\frac{1}{k}+ \frac{(-1)^{n}}{k+n}\right) = \sum_{k=1}^{n}(-1)^{k+1} \frac{(-1)^{n}k+k+n}{k(k+n)}
$$
and then consider two cases: $n$ is odd and $n$ is even.
A: 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?
A: 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}
$
