In this post, we will learn about Cauch’s criterion for the convergence of a series. The statement of Cauch’s criterion for series along with some applications will be provided here.

## Statement of Cauch’s criterion for series

**Statement:** The series $\sum_{n=1}^\infty a_n$ converges if and only if given any ε>0, there exists a positive integer N (depending upon ε) such that whenever n ≥ N we have:

|a_{n+1 }+ a_{n+2 }+ … +a_{n+p}| < ε for p=1, 2, 3, …

**Corollary:**

Put p=1 in Cauch’s criterion, so we get the following:

If the infinite series ∑_{n≥1} a_{n} converges, then for any ε>0, there is a positive integer N (depending upon ε) such that

|a_{n+1}| < ε whenever n ≥ N.

⇒ lim_{n→∞} a_{n+1 }= 0

⇒ lim_{n→∞} a_{n }= 0.

Thus, as an application of Cauch’s criterion, we deduce the theorem below.

## Application of Cauch’s criterion for series

**Theorem:** If the infinite series ∑_{n≥1} a_{n} converges, then lim_{n→∞} a_{n}= 0.

**But the converse is not true.**

That is, if lim_{n→∞} a_{n }≠ 0, then the infinite series ∑_{n≥1} a_{n} does not converge. For example,

∑_{n≥1} 1/n diverges, but we have 1/n **→** 0 when n**→**∞.

**Example 1:**

∑_{n≥1} $\dfrac{n}{n+1}$

Here, $a_n=\dfrac{n}{n+1}$ $=\dfrac{1}{1+\frac{1}{n}}$ **→** 1 as n**→**∞.

So the limit of a_{n} is not 0.

Thus, the above series does not converge.

**Example 2:**

∑_{n≥1} cos$\dfrac{x}{n}$

Here, a_{n} = cos$\dfrac{x}{n}$

Note that lim_{n→∞} cos$\dfrac{x}{n}$ =1 for any real number x.

So the limit of a_{n} is not equal to 0.

Thus, the series ∑_{n≥1} cos(x/n) does not converge.

**Example 3:**

∑_{n≥1} $\dfrac{n^n}{n!}$

Here, a_{n} = $\dfrac{n^n}{n!}$

= $\dfrac{n⋅n⋅n⋅…⋅n (n \text{ times})}{1⋅2⋅3⋅…⋅n}$ > 1 whenever n>1.

So a_{n} does not tend to 0 when n**→**∞. As a result, the series ∑_{n≥1} n^{n}/n! is not convergent.

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