I am having trouble proving the following using Mathematical Induction.

1,3,5...(2n-1)∕2,4,6...(2n) ≥ 1∕2n

I cannot seem to understand using Induction to prove a fractional expression such as the one above.

I assume the proof for the base case of n is required, like most induction questions.

However I do seem to be lost with this one.

All help is appreciated.

  • $\begingroup$ Hint: the inequality is the same as $1\times3\times5\times\cdots\times(2n-1)\ge2\times4\times6\times\cdots\times(2n-2)$, which you may find easier. $\endgroup$
    – David
    Jul 27, 2018 at 3:43

1 Answer 1


The base case is $n=1$ (as the quantities involved are not defined for $n=0$). The property then states $\frac{1}{2}\geq \frac{1}{2}$, which is true.

Assume now that the property is true for some $n\in \mathbb{N}$. We want to show that it is true for $n+1$. We have $$\prod_{k=1}^{n+1}\frac{2k-1}{2k}=\frac{2n+1}{2(n+1)}\times \prod_{k=1}^{n}\frac{2k-1}{2k}$$ where we took out the $k=n+1$-term. Now, note that the product on the right is the quantity involved in our property, at rank $n$. We may use the induction hypothesis to get

$$\prod_{k=1}^{n+1}\frac{2k-1}{2k}\geq \frac{2n+1}{2(n+1)}\times \frac{1}{2n}$$

We want this last quantity to be greater or equal $\frac{1}{2(n+1)}$. Thus, we need to show that $\frac{2n+1}{2n}\geq 1$, which actually is obvious. So we conclude

$$\prod_{k=1}^{n+1}\frac{2k-1}{2k}\geq \frac{1}{2(n+1)}$$

which exactly is the property at rank $n+1$. This concludes the proof, using the principle of induction.


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