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?

the first two terms are both bigger then $1/4.$   Next two terms, denominators 5,7, each reciprocal is bigger then $1/8$ sum is bigger then $1/4.$  Next denominators 9,11,13,15, each bigger than $1/16,$ sum is bigger than $1/4.$  Then denominators 17, 19, 21, 23, 25, 27, 29, 31, each reciprocal bigger than $1/32,$  the sum of eight terms is bigger than $1/4.$
And so on
