# For infinite series convergence/divergence: Why doesnt meeting the conditions of the Divergence test imply the Cauchy Convergence Critierion

Assume that the limit of the sequence is zero, $\lim_{n\to\infty}a_n=0$. So its not plainly obvious if the series $\sum a_n$ converges or diverges.

I have wondered for some time. If $\lim_{n\to\infty}a_n=0$ then it must be the case that $\lim_{n\to\infty}a_{n+1}=0$ and $\lim_{n\to\infty}a_{n+2}=0$ and so on.

Does it not make sense then that $\lim_{n\to\infty}a_n + a_{n+1} + a_{n+2}=0$

Cauchy's Convergence test basically says this, does it not? $$\lim_{\substack{m\to\infty\\n\to\infty} }\sum_{k=m}^{m+n}a_k = 0$$

I realize that Cauchy's Convergence Criterion is a necessary and sufficient condition for convergence and divergence, while the Divergence test is just necessary for convergence. That said, I am wondering why. Why doesnt meeting the divergence test imply meeting Cauchy's convergence criterion?

• Excellent question! – Zev Chonoles Feb 10 '13 at 19:19
• The difference is when you choose $\epsilon$ in your limits. Cauchy says for a given $\epsilon$, there exists an $N$ such that the sum $|a_n + a_{n+1} + ... + a_{n+M}| < \epsilon$ for $n > N$ and all $M > 1$. The thing you are saying is for a fixed $M$ and $\epsilon > 0$, there exists an $N$ for which $|a_n + a_{n+1} + ... + a_{n+M}| < \epsilon$. For different $M$, you will choose a different $\epsilon$ here – muzzlator Feb 10 '13 at 19:21
• For a fixed number of terms, the limit is indeed $0$, but that is not enough for convergence. – André Nicolas Feb 10 '13 at 19:23
• @muzzlator Would you like to put that in an answer? :-) – arkeet Oct 21 '16 at 23:22
• @arkeet sure, done – muzzlator Nov 2 '16 at 2:39

## 1 Answer

The difference is when you choose $\epsilon$ in your limits. Cauchy says for a given $\epsilon>0$, there exists an $M$ such that the sum $|a_n+a_{n+1}+\cdots+a_{n+M}|<\epsilon$ for $n > N$ and all $M>1$. The thing you are saying is for a fixed $M$ and $\epsilon>0$, there exists an $N$ for which $|a_n+a_{n+1}+...+a_{n+M}|<\epsilon$. For different $M$, you will choose a different $\epsilon$ here.