Inductive proof that sum of reciprocals of odd natural numbers diverges? I am trying to prove via induction that for any natural number $N$, there exists an $n$ such that:
$$
\sum_{i=1}^{n} \dfrac{1}{2i-1} > N
$$
Or
$$
1 + \dfrac{1}{3} + \dfrac{1}{5} + ... + \dfrac{1}{2n-1} > N
$$
I've started by attempting to use induction on $N$.  The basis step is easy.  Let $N=1$, and it is clear that $1 + \dfrac{1}{3} > N$, so $n = 2$ suffices to show the basis is true.
For the inductive step, I've started with this:
For some $k > 0$, show that
$$
1 + \dfrac{1}{3} + \dfrac{1}{5} + ... + \dfrac{1}{2n-1} + \dfrac{1}{2(n+1)-1} + ... + \dfrac{1}{2(n+k)-1} > N + 1 
$$
Now I've tweaked this inequality in a bunch of different ways, but I haven't been able to get closer to proving that it's true for a certain $k$.
The closest I've got is trying to show that for a given $k$,
$$
\dfrac{1}{2(n+1)-1} + ... + \dfrac{1}{2(n+k)-1} > 1
$$
Can anyone point me in the right direction?

$$
\frac{1}{2(n+1)-1} > \frac{1}{2(n+2)-1} > \cdots > \frac{1}{2(n+k)-1}.
$$
Therefore
$$
\frac{1}{2(n+1)-1} + \frac{1}{2(n+2)-1} + \cdots + \frac{1}{2(n+k)-1}
> \frac{k}{2(n+k)-1}.
$$
The right-hand side will never be greater than $1.$ But if you can make it greater than, say, $\frac14,$ all you need to do is repeat the process four times and you will have a sum of terms that is greater than $1.$
