Inductive proof that sum of reciprocals of odd natural numbers diverges?

Question

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?

Answer 1

$$ \frac{1}{2(n+1)-1} > \frac{1}{2(n+2)-1} > \cdots > \frac{1}{2(n+k)-1}. $$

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 $1.$

Answer 2

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