Define $L_n=\frac{2n-1}{2}-\frac{2n-2}{3}+\frac{2n-3}{4}-+...-\frac{2}{2n-1}+\frac{1}{2n}$ and $R_n=\frac{1}{n+1}+\frac{3}{n+2}+...+\frac{2n-1}{2n}$. We get $d_n:=(R_{n+1}-R_n)-(L_{n+1}-L_n)=1-\frac12+...-\frac1{2n}-\left(\frac1{n+1}+\frac1{n+2}+...+\frac1{2n}\right)$. By induction we see easily $d_n=0$, as the induction step from $n-1$ to $n$ is the obvious identity $\frac1{2n-1}-\frac1{2n}=-\frac1{n}+\frac1{2n-1}+\frac1{2n}$. So $R_{n+1}-L_{n+1}=R_n-L_n=...=R_1-L_1=0$ qed. I wonder if this funny identity can be derived directly without induction.

Of course it can. $\sum_{k=1}^{2n} \frac {(-1)^{k+1}} {k} = \sum_{k=1}^{2n} \frac {1} {k} - 2\sum_{k=1}^{n} \frac {1} {2k} = H_{2n} - H_n$. $\sum_{k=n+1}^{2n} \frac {1} {k} = \sum_{k=1}^{2n} \frac {1} {k} - \sum_{k=1}^{n} \frac {1} {k}= H_{2n} - H_n$. It is the very identity that is used to prove that both these forms converge to $\ln 2$.

mavropnevma wrote: Of course it can. Sure for the second identity. But based on this, I mean there is still a kind of induction for the first one.

Well, once you establish $(R_{n+1}-R_n)-(L_{n+1}-L_n)=1-\frac12+...-\frac1{2n}-\left(\frac1{n+1}+\frac1{n+2}+...+\frac1{2n}\right) = 0$, this may qualify as induction, true - but a very obvious one

It is Catalan's identity.

We shall prove the more general identity: $\frac{2n-1}{2}-\frac{2n-2}{3}+\frac{2n-3}{4}-+...-\frac{2}{2n-1}+\frac{1}{2n}=\frac{1}{n+1}+\frac{3}{n+2}+...+\frac{2n-1}{2n}$ where $n$ is a positive integer (the problem deals with $n=998$) The addends of the LHS are of the form $\frac{2n-k}{k+1}$, $k=1,2,...,2n-1$ The addends of the RHS are of the form $\frac{2l-1}{n+l}$, $l=1,2,...,n$ We have: $\frac{2n-k}{k+1}=\frac{2n+1}{k+1}-1$ $\frac{2l-1}{n+l}=2-\frac{2n+1}{n+l}$ The identity becomes: $\left(\frac{2n+1}{2}-1\right)-\left(\frac{2n+1}{3}-1\right)+...-\left(\frac{2n+1}{2n-1}-1\right)+\left(\frac{2n+1}{2n}-1\right)\\=\left(2-\frac{2n+1}{n+1}\right)+\left(2-\frac{2n+1}{n+2}\right)+\left(2-\frac{2n+1}{n+3}\right)+...+\left(2-\frac{2n+1}{2n}\right)$ Or $(2n+1)\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-...-\frac{1}{2n-1}+\frac{1}{2n}\right)-1=2n-(2n+1)\left(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}\right)$