I need help to proof the following inequality by induction
$$P(n):\quad\sum_{i=1}^{2^n} \frac{1}{2i-1} > \frac{n+3}{4}$$
Here's what I did so far.. Please tell me where I'm wrong or how to continue
First I check $P(1)$ is true, Then I asume $P(n)$ is true for all Natural numbers $n\leq k$ and want to see if it implies the following statement:
$$P(k+1): \quad\sum_{i=1}^{2^{k+1}} \frac{1}{2i-1} > \frac{(k+1)+3}{4} $$
So I begin like this
$$\sum_{i=1}^{2^{k+1}} \frac{1}{2i-1} = \sum_{i=1}^{2^{k}} \frac{1}{2i-1} + \sum_{i=2^k+1}^{2^{k+1}} \frac{1}{2i-1}$$
Then by induction hypothesis
$$\sum_{i=1}^{2^{k}} \frac{1}{2i-1} + \sum_{i=2^k+1}^{2^{k+1}} \frac{1}{2i-1} >\frac{k+3}{4}+\sum_{i=2^k+1}^{2^{k+1}} \frac{1}{2i-1}$$
Here is where I'm stuck.
I think that I should prove that $$\sum_{i=2^k+1}^{2^{k+1}} \frac{1}{2i-1} > \frac{1}{4}$$ That way, I could state that $p(k+1)$ is valid $\forall k \in N$.
Is this the right way to solve it? How should I continue?
Continuing from where you left, $$\sum_{i=2^k+1}^{2^{k+1}} \frac{1}{2i-1} > \frac{1}{4}$$ would imply $$\sum_{i=2^k+1}^{2^{k+1}} \frac{2^{k+2}}{2i-1} > \frac{2^{k+2}}{4}$$ or $$\sum_{i=2^k+1}^{2^{k+1}} \frac{2^{k+2}}{2i-1} > 2^{k}$$ which is true since for all $i$, every term on the left will be greater than one, and you'll be summing it up $2^k$ times.