How do you prove this without using induction?

Question

How do you prove this without using induction

$$\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n-1}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2n-1}$$

Answer 1

Note that we can write

$$\begin{align} \sum_{k=1}^{2n} \frac{(-1)^{k-1}}{k}&=\sum_{k=1}^{n}\left(\frac{1}{2k-1}-\frac{1}{2k}\right)\\\\ &=\sum_{k=1}^{n}\left(\frac{1}{2k-1}+\frac{1}{2k}\right)-\sum_{k=1}^{n}\frac1k\\\\ &=\sum_{k=1}^{2n}\frac1k -\sum_{k=1}^{n}\frac1k\\\\ \sum_{k=1}^{2n} \frac{(-1)^{k-1}}{k}&=\sum_{k=n+1}^{2n}\frac1k \tag 1 \end{align}$$

Thus, subtracting the last term of the sum on the left-hand side of $(1)$ from both sides yields the coveted equality

$$\sum_{k=1}^{2n-1} \frac{(-1)^{k-1}}{k}=\sum_{k=n}^{2n-1}\frac1k$$

Answer 2

$$\sum_{r=1}^{2n}\dfrac1r=\sum_{r=1}^n\dfrac1{2r-1}+\sum_{r=1}^n\dfrac1{2r} =\sum_{r=1}^n\dfrac1{2r-1}+\dfrac12\sum_{r=1}^n\dfrac1r$$

So, $$\sum_{r=n}^{2n}\dfrac1r=\sum_{r=1}^{2n}\dfrac1r-\sum_{r=1}^n\dfrac1r =\sum_{r=n}^{2n}\dfrac1r=\left(\sum_{r=1}^n\dfrac1{2r-1}+\dfrac12\sum_{r=1}^n\dfrac1r\right)-\sum_{r=1}^n\dfrac1r=?$$

Answer 3

Let $A(n)=\frac 11 + \frac 12 + \frac 13 + \dots + \frac 1n$. We have $$ \begin{align} A(2n)&=\frac 11 + \frac 12 + \frac 13 + \dots + \frac 1n + \dots + \frac 1{2n} \\ &=\left(\frac 11 + \frac 13 + \frac 15 + \dots + \frac 1{2n-1}\right)+\left(\frac 12 + \frac 14 + \frac 16 + \dots + \frac 1{2n}\right) \\ &=\left(\frac 11 + \frac 13 + \frac 15 + \dots + \frac 1{2n-1}\right)+\frac 12 A(n) \end{align} $$ However, we also have $$ \begin{align} A(2n)&=\frac 11 + \frac 12 + \frac 13 + \dots + \frac 1n + \dots + \frac 1{2n} \\ &=A(n)+\left(\frac 1{n+1} + \frac 1{n+2} +\dots + \frac 1{2n} + \dots + \frac 1{2n} \right) \end{align} $$ Thus $$ \begin{align} A(n)+\left(\frac 1{n+1} + \frac 1{n+2} +\dots + \frac 1{2n} + \dots + \frac 1{2n} \right) &=\left(\frac 11 + \frac 13 + \frac 15 + \dots + \frac 1{2n-1}\right)+\frac 12 A(n) \\ \frac 1{n+1} + \frac 1{n+2} +\dots + \frac 1{2n} + \dots + \frac 1{2n} &=\left(\frac 11 + \frac 13 + \frac 15 + \dots + \frac 1{2n-1}\right)-\frac 12 A(n) \\ &= \frac 11 -\frac 12 + \frac 15-\frac 14 + \dots + \frac 1{2n-1} \end{align} $$