how can I prove this $\sum_{k=1}^{n}\frac{1}{4k^{2}-2k}=\sum_{k=n+1}^{2n}\frac{1}{k}$

Question

How can I prove the following equation?:

$$s=\sum_{k=1}^{n}\frac{1}{4k^{2}-2k}=\sum_{k=n+1}^{2n}\frac{1}{k}$$

Simplifying both terms of the equation:

$$\sum_{k=1}^{n}\frac{1}{4k^{2}-2k} = \sum_{k=1}^{n}\frac{1}{(2k-1)2k} =(\sum_{k=1}^{n}\frac{1}{2k-1}-\sum_{k=1}^{n}\frac{1}{2k})$$

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

Now we have:

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

How can I continue?

Answer 1

A non-inductive proof , $$ S= \sum_{k=1}^{n}\frac{1}{4k^2-2k}=\frac{1}{2}\sum_{k=1}^{n}\frac{1}{(2k-1)(k)} $$ $$ S= \frac{1}{2}\left ( \sum_{k=1}^{n}\frac{2}{(2k-1)}-\frac{1}{k} \right) $$ $$S= \sum_{k=1}^{n}\frac{1}{(2k-1)}-\sum_{k=1}^{n}\frac{1}{2k} $$ Also it is easy to see that, $$ \sum_{k=1}^{2n}\frac{1}{k} = \sum_{k=1}^{n}\frac{1}{(2k-1)}+\sum_{k=1}^{n}\frac{1}{2k} $$ Subtracting the 2 equations we get, $$ S =\sum_{k=1}^{2n}\frac{1}{k} - 2\sum_{k=1}^{n}\frac{1}{2k} =\sum_{k=n+1}^{2n}\frac{1}{k} $$ Hence, proved.

Answer 2

Let's do a cute induction on $n$:

For $n = 1$, we have $\frac{1}{4 - 2} = \frac{1}{2}$, so the claim holds.

Assume $n > 1$ and the claim holds for $n - 1$. Then $$\sum_{k = 1}^n \frac{1}{4k^2 - 2k} = \frac{1}{4n^2 - 2n} + \sum_{k = 1}^{n - 1} \frac{1}{4k^2 - 2k} = \frac{1}{4n^2 - 2n} + \sum_{k = n}^{2(n -1)} \frac{1}{k}$$ by the induction hypothesis. So we only need to show that $\frac{1}{2n} + \frac{1}{2n - 1} - \frac{1}{n} = \frac{1}{4n^2 - 2n}$, which is just a direct computation.

Answer 3

The core part of the induction argument spelled out more explicitly: \begin{align} \sum_{i=1}^{k+1}\frac{1}{2i(2i-1)}&= \sum_{i=1}^k\frac{1}{2i(2i-1)}+\frac{1}{2(k+1)(2k+1)}\\[1em] &= \sum_{i=k+1}^{2k}\frac{1}{i}+\frac{1}{2(k+1)(2k+1)}\\[1em] &= \sum_{i=k+2}^{2k+2}\frac{1}{i}+\frac{1}{k+1}-\frac{1}{2k+1}-\frac{1}{2k+2}+\frac{1}{2(k+1)(2k+1)}\\[1em] &= \sum_{i=k+2}^{2k+2}\frac{1}{i}+\frac{2(2k+1)-2(k+1)-(2k+1)}{2(k+1)(2k+1)}+\frac{1}{2(k+1)(2k+1)}\\[1em] &= \sum_{i=k+2}^{2k+2}\frac{1}{i}-\frac{1}{2(k+1)(2k+1)}+\frac{1}{2(k+1)(2k+1)}\\[1em] &= \sum_{i=k+2}^{2k+2}\frac{1}{i}. \end{align}