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.