How could we show the divergence of a sequence solely by using the definition of divergence ? Also how can this general method be manipulated to work for series too?
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What definition of divergence do you mean? Usually it is defined as not convergent, so you would work with the negation of that definition. See Negating the Definition of a Convergent Sequence to Find the Definition of a Divergent Sequence – StackTD May 13 '16 at 07:52
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For example the definition of divergence to infinity, i.e. S_n tends to infinity if for all H >0 there exists n>n_o : S_n >H – iheartmaths May 13 '16 at 07:58
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Alright; in that case: I answered with an example. – StackTD May 13 '16 at 12:43
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What do you mean with the 'manipulating it for series'-part? – StackTD May 13 '16 at 12:51
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Based on OP's comment, here is a simple example to prove the sequence with terms $a_n = n^2$ tends to infinity by using the definition; we write: $$\lim_{n \to \infty} n^2= +\infty$$ if the following is satisfied: $$ \forall H > 0, \exists N \in \mathbb{N}: \color{blue}{n > N} \implies \color{red}{n^2 > H}$$ For an arbitrary $H>0$, pick $N$ so that $N > \sqrt{H}$; you then have that whenever $\color{blue}{n > N}$: $$a_n = \color{red}{n^2} > N^2 \color{red}{>} \sqrt{H}^2=\color{red}{H}$$ so $\color{red}{n^2 > H}$.
How do you arrive at a good choice for $N$? You usually work backwards. You want $n^2 > H$ to hold for any $H$ and you're looking for a condition on $n$ to make this work. Since $n^2 > H \Leftrightarrow n > \sqrt{H}$ if $H$ is positive, a good choice for $N$ is anything above $\sqrt{H}$.
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