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

2 Answers
Dec 14, 2016

As:

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

and:

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

the series is convergent.

Explanation:

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.

Dec 14, 2016

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.