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:
|an+1 + an+2 + … +an+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 an converges, then for any ε>0, there is a positive integer N (depending upon ε) such that
|an+1| < ε whenever n ≥ N.
⇒ limn→∞ an+1 = 0
⇒ limn→∞ an = 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 an converges, then limn→∞ an= 0.
But the converse is not true.
That is, if limn→∞ an ≠ 0, then the infinite series ∑n≥1 an 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 an is not 0.
Thus, the above series does not converge.
Example 2:
∑n≥1 cos$\dfrac{x}{n}$
Here, an = cos$\dfrac{x}{n}$
Note that limn→∞ cos$\dfrac{x}{n}$ =1 for any real number x.
So the limit of an 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, an = $\dfrac{n^n}{n!}$
= $\dfrac{n⋅n⋅n⋅…⋅n (n \text{ times})}{1⋅2⋅3⋅…⋅n}$ > 1 whenever n>1.
So an does not tend to 0 when n→∞. As a result, the series ∑n≥1 nn/n! is not convergent.
ALSO READ:
Rolle’s Theorem Questions-Answers