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Comparison test

It is a way to test if a given series of non-negative real numbers converges or not. The technique is to compare it with another series of non-negative real numbers whose behaviour is known.

The basic idea is like this:

Suppose that $\sum a_n$ converges, where $\forall n~~a_n\geq0.$ So we are adding more and more and more $a_n$'s (so the sum is growing and growing and growing), but still the sum is not blowing up to infinity. Clearly, this means that the $a_n$'s are getting smaller and smaller and smaller pretty fast. Now suppose that I give you another series $\sum b_n$ where $\forall n~~0\leq b_n\leq a_n.$

It should be intuitively quite obvious that $\sum b_n$ must also converge. Well, that is indeed true.

This is called comparison test.
Comparison test Let $\sum a_n$ and $\sum b_n$ be two series of non-negative real numbers such that $\forall n~~b_n\leq a_n.$ Then

Proof: Consider the sequence of partial sums $\left(\sum_1^n b_k\right)_n.$ Clearly, this is a non-decreasing sequence (since $b_n$' are all $\geq 0$). Also this is bouded from above, since $$ \forall n~~\left( \sum_1^n b_k \leq \sum_1^n na_k \leq \sum_1^\infty a_k\right). $$ Hence the sequence must converge, as required.

The second part is just the contrapositive of the first ($A\Rightarrow B$ is the same as $\neg B\Rightarrow \neg A$). [QED]

In fact, you do not even need $b_n\leq a_n$ to hold for all $n$'s. It is enough if the ineuqlity holds $\forall n\geq N$ for some $N.$