Question

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

Answer 1

The series:

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

is conditionally convergent but not absolutely convergent.

Explanation:

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

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