# Prove inequality using induction with a sequence

The sequence is defined as $$d_n = 1$$ if $$n = 0$$, $$\frac{n}{d_{n-1}}$$ otherwise. The goal is to prove $$\forall n \in \mathbb{Z^+}, d_{2n-1} \leq \sqrt{2n-1}$$ using induction. I successfully proved the base case, and laid out the induction step: assume $$d_{2k-1} \leq \sqrt{2k-1}$$ prove $$d_{2k+1} \leq \sqrt{2k+1}$$ But I am now struggling to algebraically prove the later inequality. I expressed $$d_{2k+1}$$ in terms of $$d_{2k-1}$$: $$d_{2k+1} = \frac{(2k+1)d_{2k-1}}{2k}$$ But I do not see any steps I can take from here. Any help is appreciated! I am new to this resource, please let me know if I formatted anything incorrectly!

Starting from $$d_{2k-1} \leq \sqrt{2k-1}$$ multiply both sides with $$2k+1$$ and you will get $$2k\cdot d_{2k+1} = (2k+1)d_{2k-1} \leq (2k+1)\sqrt{2k-1}$$ Divide both sides by $$2k$$:
$$d_{2k+1} \le \sqrt{2k+1} \frac{\sqrt{(2k-1)(2k+1)}}{2k} \leq \sqrt{2k+1} \sqrt{\frac{4k^2-1}{4k^2}} \le \sqrt{2k+1}$$