# How do you determine the convergence or divergence of #Sigma ((-1)^(n)n)/(n^2+1)# from #[1,oo)#?

##### 2 Answers

As:

and:

the series is convergent.

#### Explanation:

An alternative series:

converges if the succession

So we make the test:

As the denominator is always positive we can focus on the numerator:

which is always true, for

The we check that:

So both conditions are satisified and the series is convergent.

Another way that showing

#d/dx(x/(x^2+1))=((d/dxx)(x^2+1)-x(d/dx(x^2+1)))/(x^2+1)^2#

#color(white)(d/dx(x/(x^2+1)))=(1(x^2+1)-x(2x))/(x^2+1)^2#

#color(white)(d/dx(x/(x^2+1)))=(1-x^2)/(x^2+1)^2#

Examining the sign of the derivative, we see that the denominator is always positive. Thus the sign of the derivative as a whole is dependent on the sign of the numerator. When

A negative derivative shows a decreasing function. Thus,

Using this in conjunction with the fact that

#lim_(nrarroo)a_n=lim_(nrarroo)n/(n^2+1)=lim_(nrarroo)(1/n)/(1+1/n^2)=0/(1+0)=0# ,

we can claim that

We can go on to note that **conditionally convergent**.