Review question

# Can we find an inequality from the expansion of $[ 1+1/\sqrt{n}]^n$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6770

## Solution

By considering the first three terms of the binomial expansion of $\left[ 1+\left(\dfrac{1}{\sqrt{n}}\right) \right]^n$, where $n$ is an integer greater than $1$, prove that $\left[1+\left(\frac{1}{\sqrt{n}}\right)\right]^{n-2} \geq \tfrac{1}{2}n.$

Since $n \geq 2$, we know that we have at least three terms in the binomial expansion (which in general has $n+1$ terms).

Furthermore, we know every term is positive, since $1$ and $1/\sqrt{n}$ both are, which means we can truncate the sum to obtain an inequality. Hence \begin{align*} \left[ 1+\frac{1}{\sqrt{n}} \right]^n &= \sum_{k=1}^n {n \choose k}\left(\frac{1}{\sqrt{n}}\right)^k \\ &= 1+n\frac{1}{\sqrt{n}}+\frac{n(n-1)}{2}\left(\frac{1}{\sqrt{n}}\right)^2+\cdots \\ &\geq 1+n\frac{1}{\sqrt{n}}+\frac{n(n-1)}{2}\left(\frac{1}{\sqrt{n}}\right)^2\\ &= 1+\sqrt{n}+\frac{n-1}{2}. \end{align*}

This looks promising, but the inequality we are trying to prove has $n-2$ instead of $n$ in the exponent on the left hand side.

To fix this, there are (at least) two approaches we can take.

#### Approach 1: Divide through by $(1+1/\sqrt{n})^2$

The most direct approach is to simply divide both sides of our inequality by $(1+1/\sqrt{n})^2$.

After this, the left hand side is what we want, but the right hand side is a bit of a mess: $\left[ 1+\frac{1}{\sqrt{n}} \right]^{n-2} \geq \frac{1+\sqrt{n}+\frac{n-1}{2}}{\left(1+\frac{1}{\sqrt{n}}\right)^2}.$

To clean it up, we simplify the numerator and denominator of the right hand side. The numerator can be written over a common denominator to get

$1+\sqrt{n}+\frac{n-1}{2}=\frac{1+2\sqrt{n}+n}{2}= \frac{1}{2}(1+\sqrt{n})^2.$

(For the final step here, we noticed that $1+2\sqrt{n}+n$ is the square of $1+\sqrt{n}$.)

Similarly, the denominator can be expanded to get $\left(1+\frac{1}{\sqrt{n}}\right)^2= \left(\frac{\sqrt{n}+1}{\sqrt{n}}\right)^2= \frac{1}{n}\left(1+\sqrt{n}\right)^2.$

Substituting these back in we obtain

$\left[ 1+\frac{1}{\sqrt{n}} \right]^{n-2} \geq \frac{\frac{1}{2}(1+\sqrt{n})^2}{\frac{1}{n}\left(1+\sqrt{n}\right)^2}=\tfrac{1}{2}n.$

#### Approach 2: Manipulate the right hand side to find hidden factors of $(1+1/\sqrt{n})$.

In this less direct (but essentially equivalent) approach we manipulate the right hand side by looking for factors of $(1+1/\sqrt{n})$ that we can cancel from both sides.

For this it’s handy to note that we can write $n-1=(\sqrt{n}+1)(\sqrt{n}-1)$. Thus \begin{align*} \left[ 1+\frac{1}{\sqrt{n}} \right]^n &\geq 1+\sqrt{n}+\frac{n-1}{2}. \\ &= 1+\sqrt{n}+\frac{(\sqrt{n}+1)(\sqrt{n}-1)}{2}. \\ &= (1+\sqrt{n})\left(1+\frac{\sqrt{n}-1}{2}\right). \\ &= (1+\sqrt{n})\left(\frac{\sqrt{n}+1}{2}\right). \\ &= \frac{1}{2}(1+\sqrt{n})^2 \\ \end{align*} This still isn’t quite what we want — on the right hand side we have $(1+\sqrt{n})$, when we want $(1+1/\sqrt{n})$ — but we can fix this by noting that $(1+\sqrt{n})=\sqrt{n}(1+1/\sqrt{n})$. Thus \begin{align*} \left[ 1+\frac{1}{\sqrt{n}} \right]^n &\geq \frac{1}{2}(1+\sqrt{n})^2 \\ &= \frac{1}{2}n\left[1+\frac{1}{\sqrt{n}}\right]^2, \\ \end{align*}

which, after dividing through by $(1+1/\sqrt{n})^2$, implies that $\left[1+\frac{1}{\sqrt{n}}\right]^{n-2} \geq \frac{1}{2}n,$ as desired.

Is it important that $n>1$?

You might like to check if the inequality is true at $n=1$, or $n=0$. You might then consider what would happen if $n$ was allowed to take non-integer real values.

Where do issues arise?