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

An alternative series:

#sum_(n=1)^(oo) (-1)^n a_n#

converges if the succession #{a_n}# is decreasing and converges to zero.

So we make the test:

#a_(n+1) <= a_n#

#(n+1)/((n+1)^2+1) <= n/(n^2+1)#

#(n+1)/((n+1)^2+1) - n/(n^2+1) <=0#

#( (n+1)(n^2+1)-n( (n+1)^2+1)) / ( ( (n+1)^2+1)(n^2+1)) <=0#

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

#( (n+1)(n^2+1)-n( (n+1)^2+1)) = n^3+n+n^2+1 -n (n^2+2n+1+1) = n^3+n+n^2+1 -n^3-2n^2-2n = -2n^2-2n +1 <=0#

which is always true, for #n>1# so the first condition is satisfied.

The we check that:

#lim_(n->oo) a_n = lim_(n->oo) n/(n^2+1) = 0#

So both conditions are satisified and the series is convergent.

Another way that showing #a_n# is decreasing on #nin[1,oo)# other than showing that #a_(n+1)lta_n# is to find the derivative of #a_n#.

#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 #x>1#, the numerator is negative, so the derivative is negative.

A negative derivative shows a decreasing function. Thus, #a_n# is decreasing on #n in [1,oo)#.

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 #sum_(n=1)^oo((-1)^n n)/(n^2+1)# is convergent through the alternating series test.

We can go on to note that #sum_(n=1)^oon/(n^2+1)# is divergent through limit comparison with the divergent series #sum_(n=1)^oo1/n#, so we can claim that #sum_(n=1)^oo((-1)^n n)/(n^2+1)# is conditionally convergent.