If $f(n) > k$ for all $n \geq 1$, what can we say about $k$?

If $f(n) > k$ for all $n \geq 1$, what can we say about $k$?

Question

[Choose the one correct answer and explain your reasoning.]

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. $k < 1300$,

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1. $k^2<101$,

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

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1. $k < 5150$.

Suggestion

What happens when $n=1$?

What happens to the LHS when $n=2$? Is it bigger or smaller than for the case $n=1$?

Solution

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

$$\begin{align*} &(1+1)+(1+2)+(1+3)+(1+4)+\cdots +(1+100) > k\\ \iff\quad& 2+3+4+5+\cdots +101 > k\\ \iff\quad& \frac{100}{2}(2+101) > k\\ \iff\quad& 5150 > k. \end{align*}$$

Hence the answer is (d).

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