Proving inequalities with fractions: $(\frac{1}{2^n+1}+\frac{1}{2^n+2}+\frac{1}{2^n+3}+\ldots+\frac{1}{2^{n+1}})>\frac{1}{2}$

Question

Having an immense amount of trouble trying to prove the general case for inequalities involving fractions. For instance the proof is :Prove for all natural numbers,

$(\frac{1}{2^n+1}+\frac{1}{2^n+2}+\frac{1}{2^n+3}+\ldots+\frac{1}{2^{n+1}})>\frac{1}{2}$

I understand the pattern after a little help but I do not know how to finish the proof.

I am having difficulty writing a summation for the general pattern of repeating fractions and therefore using correct notation to prove the result. How should I go about doing this?

For $n=1$

$\frac{1}{3}+\frac{1}{4}>\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$

For $n=2$

$\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}>\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{1}{2}$

Answer 1

Your work gives: $$\frac{1}{2^n+1}+\frac{1}{2^n+2}+\frac{1}{2^n+3}+\ldots+\frac{1}{2^{n+1}}>$$ $$>\frac{1}{2^{n+1}}+\frac{1}{2^{n+1}}+\frac{1}{2^{n+1}}+\ldots+\frac{1}{2^{n+1}}=2^n\cdot\frac{1}{2^{n+1}}=\frac{1}{2}$$

Answer 2

This is an example $$\dfrac{1}{2^n+1}+\dfrac{1}{2^n+2}+\dfrac{1}{2^n+3}+\cdots+\dfrac{1}{2^{n+1}}\\>\dfrac{1}{2^{n+1}}+\dfrac{1}{2^{n+1}}+\cdots+ \dfrac{1}{2^{n+1}}\\=\dfrac{2^n}{2^{n+1}}=\dfrac{1}{2}$$

Answer 3

I find it often easier to express a result that uses ellipses ("...") in terms of explicit summation (or product) signs.

In your case, you want to prove $\sum_{k=1}^{2^n} \dfrac1{2^n+k} \gt \dfrac12 $.

Since $k \le 2^n$ for $1 \le k \le 2^n$ and $k \lt 2^n$ for $1 \le k \lt 2^n$, $2^n+k \le 2^{n+1}$ for $1 \le k \le 2^n$ and $2^n+k \lt 2^{n+1}$ for $1 \le k \lt 2^n$, so $\dfrac1{2^n+k} \ge \dfrac1{2^{n+1}}$ for $1 \le k \le 2^n$ and $\dfrac1{2^n+k} \gt \dfrac1{2^{n+1}}$ for $1 \le k \lt 2^n$.

Therefore $\sum_{k=1}^{2^n} \dfrac1{2^n+k} \gt \sum_{k=1}^{2^n} \dfrac1{2^{n+1}} = \dfrac{2^n}{2^{n+1}} =\dfrac12 $.