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I am having trouble understanding the basics of proving theorems by binding the term and splitting the sum. Below is a simple theorem that I am trying to prove with the bind/split technique...

a simple theorem

I seriously have no clue where to start. I understand how to use induction and whatnot but again, I'm seriously lost on this. Any dummy-level help would be greatly appreciated.

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    I think by "bind" you might mean "bound". To bound something is to supply bounds for it. Is that what you mean? – 2017-01-12
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    "Binding the term" is a method for finding the lower or upper bounds of a sum, from what I know. But yes, I need to find the bounds for this term. – 2017-01-12
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    @JoeBalin Thanks for asking this question. I am also taking Chris Marriott's algorithms course :) – 2018-04-06

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First notice that $$1^5 + \dots + n^5 = \sum_{i=1}^n i^5 = \frac{2n^6 + 6n^5 + 5n^4 -n^2}{12}.$$ Now recall the definition of $g(n) \in \Omega(f(n)).$

Definition. A function $g(n) \in \Omega(f(n))$ if $\exists n_0 \in \mathbb{N},c\in\mathbb{R}$ such that if $n \geq n_0$ then $$g(n) \geq cf(n).$$

Now, $$g(n) = \frac{2n^6 + 6n^5 + 5n^4 -n^2}{12} \geq \frac{1}{6}n^6$$ for all $n \in \mathbb{N},$ so take $n_1 = 1$ and $c = \frac{1}{6}$. Likewise we can prove that $g(n) \in \mathcal{O}(n^6)$ (exercise) and we can conclude that $g(n) \in \Theta(n^6)$.

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    Can you be more specific about which of these steps is considered "binding the term" or "splitting the sum"? – 2018-04-06
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I'd simply say that $i^5 \leq n^5$ for all $1 \leq i \leq n$, and so $$\sum_{i=1}^n i^5 \leq \sum_{i=1}^n n^5 = n^6$$

Therefore the ratio $$\frac{\sum_{i=1}^n i^5}{n^6}$$ is always at most $1$ as $n \to \infty$.