Identify whether the infinite series converge absolutely conditionally or dont sum_(n=1)^oo (-1)^(n+1) (n (arctan(n+1)-arctan (n))n=1(1)n+1(n(arctan(n+1)arctan(n)) (Apply Mean Value theorem to conclude)?

1 Answer
Jul 25, 2018

The series:

sum_(n=1)^oo (-1)^(n+1) n(arctan(n+1)-arctan (n))n=1(1)n+1n(arctan(n+1)arctan(n))

is conditionally convergent but not absolutely convergent.

Explanation:

Let a_n = n(arctan(n+1)-arctan (n))an=n(arctan(n+1)arctan(n)).

The Mean Value Theorem states that if f(x)f(x) is continuous in [a,b][a,b] and differentiable in (a,b)(a,b), then there is a point xi in (a,b)ξ(a,b) such that:

f'(c) = (f(b)-f(a))/(b-a)

Let f(x) = arctanx and a=n, b=n+1, then:

1/(1+xi^2) = arctan(n+1)-arctan(n)

with n < xi < n+1. As for x > 0 the function f'(x) = 1/(1+x^2) is strictly decreasing, we have:

1/(1+(n+1)^2) < arctan(n+1)-arctan(n) < 1/(1+n^2)

and multiplying by n:

(1) " "n/(1+(n+1)^2) < a_n < n/(1+n^2)

Now the series:

sum n/(1+(n+1)^2)

is not convergent as we can see using the limit comparison test with the harmonic series. In fact:

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

By direct comparison we can then conclude that also:

sum a_n

is not convergent hence the series in analysis is not absolutely convergent.

On the other hand, let:

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

c_n = n/(1+n^2)

We have that:

b_n >0 and c_n >0

lim_(n->oo) b_n = lim_(n->oo) c_n = 0

and:

b_n = g(n) with g(x) = x/(1+(1+x)^2)

g(x) = x/(2+2x+x^2)

(dg)/dx = ( 2+2x+x^2 -x(2+2x))/(2+2x+x^2)^2

(dg)/dx = -(3x^2-2) /(2+2x+x^2)^2 < 0 for x >1

hence: b_(n+1) < b_n and similarly:

c_n = h(n) with h(x) = x/(1+x^2)

(dh)/dx = (1+x^2-2x^2)/(1+x^2)^2

(dh)/dx = -(x^2-1)/(1+x^2)^2 < 0 for x>1

hence c_(n+1) < c_n

Based on Leibniz' theorem then the series:

sum_(n=1)^oo (-1)^(n+1)b_n

sum_(n=1)^oo (-1)^(n+1)c_n

are both convergent, then by the squeeze theorem also:

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

is convergent.