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

$(\frac{1}{2^n+1}+\frac{1}{2^n+2}+\frac{1}{2^n+3}+\ldots+\frac{1}{2^{n+1}})&gt;\frac{1}{2}$

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 $n=1$

$\frac{1}{3}+\frac{1}{4}&gt;\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$

For $n=2$

$\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}&gt;\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{1}{2}$
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
$.
Your work gives:
$$\frac{1}{2^n+1}+\frac{1}{2^n+2}+\frac{1}{2^n+3}+\ldots+\frac{1}{2^{n+1}}&gt;$$
$$&gt;\frac{1}{2^{n+1}}+\frac{1}{2^{n+1}}+\frac{1}{2^{n+1}}+\ldots+\frac{1}{2^{n+1}}=2^n\cdot\frac{1}{2^{n+1}}=\frac{1}{2}$$