Proof by induction of summation inequality: $1 + 1/2+ 1/3+ 1/4+1/5+⋯+ 1/2^n \leq n + 1$ I have been working on this problem for literally hours, and I can't come up with anything. Please help. I feel like I am about to go insane.
For all n $\in$ N, we have $$1 + \frac{1}{2}+  \frac{1}{3}+  \frac{1}{4}+\frac{1}{5} +⋯+  \frac{1}{2^n}   ≤ n + 1$$
I know that I am supposed to use a proof by induction. Here is progress so far:
1) Let P(n) be $$\sum_{i=0}^{2^n} \frac{1}{i} $$
2) Base case: $n = 1$
$$\sum_{i=1}^{2^n} \frac{1}{i} = \frac{1}{1}+ \frac{1}{2} =  \frac{3}{2}, \frac{3}{2} ≤ 2 $$
So P(1) is true.
3) Inductive hypothesis:
Suppose that P(k) is true for an arbitrary integer k $\geq$ 1
4) Inductive step:
We want to prove that P(k + 1) is true or, $$\sum_{i=1}^{2^{k+1}} \frac{1}{i} ≤ k + 2$$
By inductive hypothesis, 
$$\sum_{i=1}^{2^{k+1}} \frac{1}{i} = \sum_{i=1}^{2^k} \frac{1}{i} + \sum_{i=2^k+1}^{2^{k+1}}\frac{1}{i} ≤ k + 1 + \sum_{i=2^k+1}^{2^{k+1}}\frac{1}{i}$$
I know that I'm supposed to split the expression into two summations, but now I am completely stuck and don't know what to do from here. I got one hint that the fact $\frac{a}{b + c} < \frac{a}{b}$ is relevant, but I don't know how to get there from here. 

If $2^k + 1 \le i \le 2^{k+1}$ then $\frac 1i < \frac 1{2^k}$
So $\sum_{i=2^k+1}^{2^{k+1}}\frac{1}{i}< \sum_{i=2^k+1}^{2^{k+1}}\frac{1}{2^k} =[2^{k+1} - ([2^k + 1]-1)\times \frac 1{2^k} =   2^k\times \frac 1{2^k} = 1$
