How does the sum $ \frac{1}{2^k+1} + \cdots +\frac{1}{2^k+2^k}$ become $ \frac{2^k}{2^k+2^k}$?

Question

How does the summation $\displaystyle \frac{1}{2^k+1} + \cdots + \frac{1}{2^k+2^k}$ become $\displaystyle \frac{2^k}{2^k+2^k}$?

I understand that it is the number of terms times the smallest term, but why does this work?

Answer 1

The key observation here is that for each $i$ in $1,2,\ldots,2^k$, $$ \frac{1}{2^k+i}\geq\frac{1}{2^k+2^k}. $$ So, you have a sum of $2^k$ terms, each of which is at least $\frac{1}{2^k+2^k}$; therefore $$ \frac{1}{2^k+1}+\frac{1}{2^k+2}+\cdots+\frac{1}{2^k+2^k}\geq\underbrace{\frac{1}{2^k+2^k}+\cdots+\frac{1}{2^k+2^k}}_{2^k\text{ terms}}=\frac{2^k}{2^k+2^k}. $$

Answer 2

You replace 2^k + 1, 2^k + 2, ..., 2^k + 2^k by 2^k + 2^k and there are 2^k of them.

Answer 3

Because the denominators on the left side are $\le{2^k+2^k}$ , the sum on the left side is $\ge\frac{1}{2^k+2^k} + ... + \frac{1}{2^k+2^k}$ with $2^k$ terms and therefore the divergence of the harmonic series is proven.

Answer 4

This is actually an inequality saying $$H_{2^k} + \frac{1}{2^k+1} + \frac{1}{2^k+2} + \cdots + \frac{1}{2^k+2^k} \ge \left(1+\frac{k}{2}\right) + \frac{2^k}{2^k+2^k}$$

From what I infer, $H_{2^k} \ge 1+\frac{k}{2}$. As for the remaining terms, because each and every term between $\displaystyle \frac{1}{2^k+1}$ and $\displaystyle \frac{1}{2^k+2^k}$ is greater than or equal to $\displaystyle \frac{1}{2^k+2^k}$ and because there are $2^k$ terms, it is true that $$\frac{1}{2^k+1} + \frac{1}{2^k+2} + \cdots + \frac{1}{2^k+2^k} \ge \frac{2^k}{2^k+2^k}$$