Inequality of Finite Harmonic Series I'm asked to prove that for $n\in \mathbb{N}$
$$\frac{1}{1} + \frac{1}{2} +\cdots+\frac{1}{n} \geq 1 + \frac{n}{2}$$
by induction.
I've got a feeling that the problem isn't right (since it isn't true for any $n\in \mathbb{N}$), does anyone know the result and what it should be?
Edit: The author says that the result can be used to prove that the harmonic series diverges.
 A: It has been suggested that the inequality be reversed. Certainly we get a correct inequality, unfortunately a fairly uninteresting one. We propose that instead we let
$$f(n)=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2^n},$$
and show that $f(n)\ge 1+\dfrac{n}{2}$ 
for every integer $n \ge 0$.  
It is clear that the result holds when $n=0$. Suppose that the inequality holds at $n=k$. We show that it holds at $n=k+1$.  We have
$$f(k+1)=f(k)+\frac{1}{2^k+1}+\frac{1}{2^k+2}+\cdots +\frac{1}{2^{k+1}}.\tag{$1$}$$
Each of $\frac{1}{2^k+1}$, $\frac{1}{2^k+2}$, and so on up to $\frac{1}{2^{k+1}}$
is $\ge \frac{1}{2^{k+1}}$, and there are $2^k$ such terms. So their sum is $\ge \frac{1}{2}$. 
Thus 
$$f(k+1)\ge f(k)+\frac{1}{2}\ge 1+\frac{k}{2}+\frac{1}{2}=1+\frac{k+1}{2},$$
and now we have done the induction step.
Remark: This is a somewhat formalized version of the usual argument that the harmonic series diverges.
A: Given that the result is supposed to be usable in proving the divergence of the harmonic series, it was probably supposed to be
$$\sum_{k=1}^{2^n}\frac1k=1+\frac12+\ldots+\frac1{2^n}\ge 1+\frac{n}2\;.$$
This can indeed be proved fairly easily by induction on $n$ and can be used to prove the divergence of the harmonic series.
(My earlier answer $-$ that the inequality is the wrong way round $-$ is correct in the sense that the reversed inequality is true, but it’s clearly not what the author had in mind.)
A: Hint: In the following sum, there are $2^{k-1}$ terms, each at least as big as $2^{-k}$; therefore,
$$
\sum_{j=2^{k-1}+1}^{2^k}\frac1j\ge2^{k-1}2^{-k}=\frac12
$$
Furthermore, we also have that
$$
\begin{align}
\sum_{j=1}^{2^n}\frac1j
&=1+\sum_{k=1}^n\sum_{j=2^{k-1}+1}^{2^k}\frac1j
\end{align}
$$
