Question

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

Answer 1

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.

Answer 2

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.