<s>[INST] Prove by induction $1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{2^n} \ge 1 + \frac {n}{2}$ Prove by induction $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots + \frac{1}{2^n} \ge 1 + \frac {n}{2}$
I can't explain in words how the left hand side of the equation is achieved soI shall provide an example.
 When $n = 2$, $\frac{1}{2^2} = \frac{1}{4}$ so the left hand side will be $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}$. 
When $n = 3$, $\frac{1}{2^3} = \frac{1}{8}$, so the left hand side will be $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}$ and so forth. 
I have gotten up to the point where it is calculating the algebra. However, the result ends up being $ 1 + \frac{[2^(k+1)*k] + 2}{2(2^{k+1}} \le 1+ \frac{k+1}{2}$ when the following should be $\ge$. What can I do to solve the question? 
 [/INST] Hint: Look at your example, appropriately grouped and then with terms made systematically smaller:
$$1+{1\over2}+\left({1\over3}+{1\over4}\right)+\left({1\over5}+{1\over6}+{1\over7}+{1\over8}\right)
\ge1+{1\over2}+\left({1\over4}+{1\over4}\right)+\left({1\over8}+{1\over8}+{1\over8}+{1\over8}\right)$$
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