Prove by induction that $\sum_{i=1}^n i \geq \frac{n^2}{2}$

Question

Can someone show me a formal proof of this exercise ? \begin{equation} \sum\limits_{i=1}^n i \geq \frac{n^2}{2}, \quad \forall n \in \mathbb{N}. \end{equation}

Thanks to anyone who can help! :)

Answer 1

Hint: $\sum_{i =1}^n i \ge \frac {n^2} 2$ iff $2\sum_{i =1}^n i = (1 + .... + n) + (1 + ..... + n) \ge n^2$.

What is $(1 + ...... + n) + (1 + ...... +n)$?

Hint: addition is commutative and associative.

Answer 2

We need some proposition depending on $n$ that we can prove by induction. In this case, it is $$P(n)\ \colon\ \sum_{i=1}^n i \geq \frac{n^2}{2}.$$

First check the base case:

$$P(1)\ \colon\ 1\geq \frac{1}{2}.$$

This is true, so we have proved the base case $P(1)$. Next suppose that $P(n)$ is true, for some $n\geq 1$. We wish to prove that then $P(n+1)$ is also true, i.e. that the claim $$P(n+1)\ \colon\ \sum_{i=1}^{n+1} i \geq \frac{(n+1)^2}{2}$$ is true. Since you have assumed that $P(n)$ is true, then $$\sum_{i=1}^{n+1} i = \sum_{i=1}^n i + (n+1) \geq \frac{n^2}{2} + (n+1) = \frac{n^2+2n+2}{2}\geq \frac{n^2+2n+1}{2} = \frac{(n+1)^2}{2}.$$ This proves the induction hypothesis.