# Thread: analysis - 2

1. ## analysis - 2

How does the comparison test incorporate the cauchy criterion for convergence?

2. Originally Posted by pila0688
How does the comparison test incorporate the cauchy criterion for convergence?
Let $\displaystyle T_n=\sum_{n=1}^{N} c_n$

So the comparison test says that if $\displaystyle |a_n|\leqslant c_n~{\color{red}\star}$ for all $\displaystyle N\leqslant n$ with $\displaystyle N\in\mathbb{N}$ and$\displaystyle \sum c_n$ converges, then so does $\displaystyle \sum a_n$

Proof: $\displaystyle \sum c_n$ converges implies that $\displaystyle T_n$ converges. Now since every convergent sequence is Cauchy $\displaystyle \color{blue}\star$ it also follows that there exists some $\displaystyle K$ that if $\displaystyle K\leqslant N\leqslant M$ then $\displaystyle \left|T_M-T_N\right|<\varepsilon$. But because of how we defined $\displaystyle T_N$ we see this is equivalent to $\displaystyle \sum_{n=N}^{M}c_n<\varepsilon$. But by $\displaystyle \color{red}\star$

$\displaystyle \sum_{n=N}^{M}|a_n|\leqslant\sum_{n=N}^{M}c_n<\var epsilon$

And

$\displaystyle \left|\sum_{n=N}^{M}a_n\right|\leqslant\sum_{n=N}^ {M}|a_n|\leqslant\sum_{n=N}^{M}c_n<\varepsilon$.

From there we conclude that $\displaystyle \sum a_n$ converges

It is clear that we used the Cauchy convergence criterion at $\displaystyle \color{blue}\star$