Review question

# If $f(n) > k$ for all $n \geq 1$, what can we say about $k$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7031

## Solution

The inequality $(n+1) + \bigl(n^4+2\bigr) + \bigl(n^9+3\bigr) + \bigl(n^{16}+4\bigr) + \cdots + \bigl(n^{10000}+100\bigr) > k$ is true for all $n\ge 1$. It follows that

1. $k < 1300$,

2. $k^2<101$,

3. $k \ge 101^{10000}$,

4. $k < 5150$.

To make this a lot simpler, note that the expression $(n+1) + \bigl(n^4+2\bigr) + \bigl(n^9+3\bigr) + \bigl(n^{16}+4\bigr) + \cdots + \bigl(n^{10000}+100\bigr)$ increases as $n$ increases, so the inequality holds for all $n \ge 1$ if it holds for $n = 1$. So we need \begin{align*} &(1+1)+(1+2)+(1+3)+(1+4)+\cdots +(1+100) > k\\ \iff\quad& 2+3+4+5+\cdots +101 > k\\ \iff\quad& \frac{100}{2}(2+101) > k\\ \iff\quad& 5150 > k. \end{align*}

Alternatively, we could have broken up the sum here as $(1+1+\cdots+1)+(1+2+\cdots+100)=100+\tfrac{1}{2}\times100\times101.$