Prove by induction that $\sum_{k=n+1}^{2n} \frac{1}{k} = \sum_{m=1}^{2n} \frac{(-1)^{m+1}}{m}, \qquad \forall n \in N$

Question

Prove by mathematical inductin that: $$\sum_{k=n+1}^{2n} \frac{1}{k} = \sum_{m=1}^{2n} \frac{(-1)^{m+1}}{m}, \qquad \forall n \in N$$ is true.

For $n=1$, ($\frac{1}{2} = \frac{1}{2})$ it holds. But what to do more? Can you give me a hint? I tried from hypothesis (when $n=j$) getting to verify that it holds for $n = j+1$, but with no result.

Thanks.

Answer 1

\begin{align}\sum_{k=(n+1)+1}^{2(n+1)}\frac1k&=\sum_{k=n+2}^{2n+2}\frac1k=-\frac1{n+1}+\frac{1}{2n+1}+\frac{1}{2n+2}+\sum_{k=n+1}^{2n}\frac1k=^\text{Ind. hypothesis}\\[0.2cm]&=-\frac1{n+1}+\frac{1}{2n+1}+\frac{1}{2n+2}+\sum_{m=1}^{2n}\frac{(-1)^{m+1}}{m}\\[0.2cm]&\overset{(*)}=-\frac1{n+1}+\frac{(-1)^{2n+2}}{2n+1}+\frac{(-1)^{2n+3}}{2n+2}+\frac{2}{2n+2}+\sum_{m=1}^{2n}\frac{(-1)^{m+1}}{m}\\[0.2cm]&=-\frac{1}{n+1}+\frac{1}{n+1}+\sum_{m=1}^{2n+2}\frac{(-1)^{m+1}}{m}=\sum_{m=1}^{2(n+1)}\frac{(-1)^{m+1}}{m} \end{align} The equation $(*)$ holds because $$\frac{1}{2n+1}=\frac{(-1)^{2n+1+1}}{2n+1}$$ and $$\frac{1}{2n+2}=-\frac{1}{2n+2}+\frac{2}{2n+2}=\frac{(-1)^{2n+2+1}}{2n+2}+\frac{1}{n+1}$$ Of course $2n+2$ is even and $2n+3$ is odd for all $n\in \mathbb N$ therefore $(-1)^{2n+2}=1$ and $(-1)^{2n+3}=-1$.

Answer 2

For $n=j$, we have $$\sum_{k=j+1}^{2j} \dfrac1k = \sum_{m=1}^{2j} \dfrac{(-1)^{m+1}}{m} \,\, (\spadesuit)$$ Hence, for $n=j+1$, we need to prove that $$\sum_{k=j+2}^{2j+2} \dfrac1k = \sum_{m=1}^{2j+2} \dfrac{(-1)^{m+1}}{m}$$ Equivalently making use of $(\spadesuit)$, we need to prove that $$\dfrac1{2j+1} + \dfrac1{2j+2} - \dfrac1{j+1} = \dfrac{(-1)^{2j+2}}{2j+1} + \dfrac{(-1)^{2j+3}}{2j+2}$$ which is trivially true.

Answer 3

Hint:

$$\sum_{k=n+2}^{2n+2}\frac{1}{k}-\sum_{m=1}^{2n+2}\frac{\left(-1\right)^{m+1}}{m}=\left[\sum_{k=n+1}^{2n}\frac{1}{k}+\left(\frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n+1}\right)\right]-\left[\sum_{m=1}^{2n}\frac{\left(-1\right)^{m+1}}{m}+\left(\frac{\left(-1\right)^{2n+2}}{2n+1}+\frac{\left(-1\right)^{2n+3}}{2n+2}\right)\right]$$

Work this out to observe that it equals $0$ if: $$\sum_{k=n+1}^{2n}\frac{1}{k}-\sum_{m=1}^{2n}\frac{\left(-1\right)^{m+1}}{m}=0$$ wich is the induction hypothese.