# analysis - 2

• November 12th 2008, 08:14 PM
pila0688
analysis - 2
How does the comparison test incorporate the cauchy criterion for convergence?
• December 13th 2008, 02:01 PM
Mathstud28
Quote:

Originally Posted by pila0688
How does the comparison test incorporate the cauchy criterion for convergence?

Let $T_n=\sum_{n=1}^{N} c_n$

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

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

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

And

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

From there we conclude that $\sum a_n$ converges

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