why is that $H_{k-1/2}-H_k = 2H_{2k}-H_k$?

Asked May 15 '22 at 04:14

Active May 15 '22 at 04:14

Viewed 64 times

In "Concrete Mathematics", question number 7.10 involves the factor $H_{k-1/2}-H_k$, which when I look at the answer, it states that the identities is valid: $$ \begin{align} H_{k-1/2}-H_k &= \frac{2}{2k-1}+\cdots+\frac{2}{1} \\ &= 2H_{2k}-H_k \end{align} $$ which the first line comes from: $$ \begin{align} H_{k-1/2} - H_{-1/2} &= \frac{1}{k-1/2}+\cdots+\frac{1}{1/2} \\ &= \frac{2}{2k-1}+\cdots+\frac{2}{1} \\ \end{align} $$

But, isn't $\frac{2}{2k-1}+\cdots+\frac{2}{1} = 2\left(\frac{1}{2k-1}+\cdots+1\right)=2H_{2k-1}$?

Apparently, $2H_{2k}-H_k$ is not equal to $2H_{2k-1}$. I can't find out where I have done wrong.

asked May 15 '22 at 04:14

Bi Ao

1

You might become less confused if you write summations instead of relying on $\cdots$. – Misha Lavrov May 15 '22 at 04:16
I think the what you are thinking as $H_{2k-1}$ is not that. It only contains odd numbers. You can add all the even terms and subtract them and manipulate to get final answer. – sku May 15 '22 at 04:33
$H_{n-1/2}=2H_{2n}-H_n-2\ln(2)$ – Ali Shadhar May 15 '22 at 18:34

why is that $H_{k-1/2}-H_k = 2H_{2k}-H_k$?

0 Answers0