Problem with proof by induction

Question

I am struggling with the following equation, which I need to proof by induction:

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

$n\in \mathbb{N}$.
I tried a few times and always got stuck.

Help would be appreciated.

Answer 1

Base case of condition is quite clearly satisfied. Now assume that the result holds for some n. We have to show that the result holds for n+1.

$$\sum_{k=1}^{2n}\frac{(-1)^{k+1}}{k}= \sum_{k=n+1}^{2n}\frac{1}{k}$$ The above expression is our assumption and we have to show, $$\sum_{k=1}^{2n+2}\frac{(-1)^{k+1}}{k}= \sum_{k=n+2}^{2n+2}\frac{1}{k}$$

So, consider the LHS of the above equation $$\sum_{k=1}^{2n+2}\frac{(-1)^{k+1}}{k}= \sum_{k=1}^{2n}\frac{(-1)^{k+1}}{k}+\frac{1}{2n+1}+\frac{-1}{2n+2}$$ $$\sum_{k=1}^{2n+2}\frac{(-1)^{k+1}}{k}= \sum_{k=n+1}^{2n}\frac{1}{k}+\frac{1}{2n+1}+\frac{-1}{2n+2}$$ $$\sum_{k=1}^{2n+2}\frac{(-1)^{k+1}}{k}= \sum_{k=n+2}^{2n}\frac{1}{k}+\frac{1}{n+1}+\frac{1}{2n+1}+\frac{-1}{2n+2}$$ $$\sum_{k=1}^{2n+2}\frac{(-1)^{k+1}}{k}= \sum_{k=n+2}^{2n+2}\frac{1}{k}$$

Hence proved.

Answer 2

The proof is simple. By induction hypothesis, assume that the statement is true for $s < n$. Then, we have

\begin{align} \sum\limits_{k = 1}^{2n} \dfrac{\left( 1 \right)^{k + 1}}{k} &= \sum\limits_{k = 1}^{2 \left( n - 1 \right)} \dfrac{\left( 1 \right)^{k + 1}}{k} + \dfrac{1}{2n - 1} - \dfrac{1}{2n} \\ &= \sum\limits_{k = n}^{2 \left( n - 1 \right)} \dfrac{1}{k} + \dfrac{1}{2n - 1} - \dfrac{1}{2n} \\ &= \sum\limits_{k = n + 1}^{2 \left( n - 1 \right)} \dfrac{1}{k} + \dfrac{1}{2n - 1} + \dfrac{1}{n} - \dfrac{1}{2n} \\ &= \sum\limits_{k = n + 1}^{2n} \dfrac{1}{k} \end{align}

which completes the proof!