The actual conditional convergence is often a alternating series test to the convergence.  In math concepts, a series or integral is considered conditionally convergent in the event it converges, but very easy converge absolutely. The best examples of conditionally convergent series would be the alternating series.

A typical conditionally convergent important is that about the non-negative real axis associated with Sin (x$^2$). An alternating series is considered conditionally convergent in the event itâ€™s convergent since it is but would certainly become divergent in the event all its terms were made optimistic. And any these kinds of absolutely convergent series can also be automatically convergent since it is. In this page you'll explore more about the concept below!

## Definition

A series $\sum$u$_{n}$ is said to be absolutely convergent if the series $\sum$u$_{n}$ obtained on taking every term of the given series with a positive sign is convergent.
i.e., if the series $\sum$ |u$_{n}$| is convergent.
Conditional Convergence: A series which is convergent but is not absolutely convergent is called a conditionally convergent series.Examples:

Example 1: The series 1 - $\frac{1}{2^{3}}$ + $\frac{1}{3^{3}}$ - $\frac{1}{4^{3}}$ + .............

is absolutely convergent because the series

1 + $\frac{1}{2^{3}}$ + $\frac{1}{3^{3}}$ +$\frac{1}{4^{3}}$ + .............
obtained on taking every term of the given series with a positive sign is convergent.

Examples 2: The series 1 - $\frac{1}{2}$ + $\frac{1}{3}$ - $\frac{1}{4}$ + .............
is convergent by leibnitz test, but the series

1 + $\frac{1}{2}$ +$\frac{1}{3}$ + $\frac{1}{4}$ + .............

obtained on taking every term with a positive sign is divergent.

Thus the given series is conditionally convergent.

## Theorem

Given below is a theorem on conditional convergence:

Statement: Every absolutely convergent series is convergent.

Proof: Let $\sum u_{n}$ be absolutely convergent, so that $\sum |u_{n}$| is convergent.
Therefore for any $\in$ > 0 , by Cauchy's General Principle of convergence. there exists a positive integer m such that

|u$_{n+1}| + |u_{n+2}| + ............ + |u_{n+p}| < \in, \forall n \geq m \wedge p \geq$ 1
Also for all n and p > 1,

|u$_{n+1} + u_{n+2} + ........+ u_{n + p}| \leq |u_{n+1}| + |u_{n+2}| + ...... +|u_{n+p}| < \in, \forall n \geq m \wedge p \geq$ 1

Hence, by cauchy's General Principle of convergence the series $\sum u_{n}$ converges.

## Examples

Examples on conditional convergence are explained below.

Example 1 : Show that the series
x + $\frac{x^{2}}{2!}$$\frac{x^{3}}{3!}$ + ........

converges absolutely for all values of x.

Solution: Let u$_{n} =$\frac{x^{n}}{n!}$Now lim$_{n->\infty}\frac{|u_{n}|}{|u_{n+1}|} = lim_{n->\infty}\frac{n+1}{|x|} \rightarrow \infty$except when x = 0. So by ratio test, the series converges absolutely for all x except possibly zero. But for x = 0 the series evidently converges absolutely. Hence, the series converges absolutely for all values of x. Example 2: Show that lim$_{n->\infty}\frac{m(m-1).......(m-n+1)}{(n-1)!}x^{n}$= 0 where |x| < 1 and m is any real number. Solution: Consider the series$\sum u_{n}$, where u$_{n}$=$\frac{m(m-1).........(m-n+1)}{(n-1)!}x^{n}$Now lim$_{n->\infty}\frac{|u_{n}|}{|u_{n+1}|}$= lim_{n->\infty}|$ $\frac{n}{m-n}$|.$\frac{1}{|x|}$

= lim$_{n->\infty}$|$\frac{1}{\frac{m}{n}- 1}$|.$\frac{1}{|x|}$ = $\frac{1}{|x|}$

Hence the series $\sum u_{n}$ converges absolutely for |x| < 1

the series $\sum u_{n}$ converges for |x| < 1

lim$_{n->\infty} u_{n}$ = 0, for |x| < 1

lim$_{n->\infty}$ $\frac{m(m-1)$.......$(m-n+1)}{(n-1)!}x^{n}$ = 0, if |x|< 1

Example 3: Show that for any fixed value of x, the series $\sum_{n = 1}^{\infty}$ $\frac{sin\ nx}{n^{2}}$ is convergent.

Solution:

Let u$_{n}$  = $\frac{sin\ nx}{n^{2}}$, so that $|u_{n}$| = $\frac{|sin\ nx|}{n^{2}}$

Now |$\frac{sin\ nx}{n^{2}}$| $\leq$ $\frac{1}{n^{2}}$, $\forall$ n and $\sum$ $\frac{1}{n^{2}}$ converges.

Hence, by comparison test, the series $\sum$ |$\frac{sin\ nx}{n^{2}}$| converges.

Since every absolutely convergent series is convergent, therefore $\sum \frac{sin\ nx}{n^{2}}$ is convergent.