How do you test for convergence for #sum((-1)^n)*(sqrt(n))*(sin(1/n))# for n is 1 to infinity?

2 Answers
Jun 25, 2015

Answer:

The series converges conditionally.

Explanation:

We have #sum_(n=1)^(+oo)(-1)^n * sqrt(n) * sin(1/n)#.

Let's write it as #sum_(n=1)^(+oo)(-1)^n * u_n#,

where #u_n = sqrt(n) * sin(1/n)#.

An alternating series is conditionally convergent if #u_n# is decreasing (so #u_(n+1)<=u_n#) for all #n \gt N#, #N \in NN#, and if #lim_(n->+oo)u_n = 0#.

Let's prove #u_(n+1)<=u_n# for all #n# big enough

Consider the function #f(x) = xsin(1/x)#, #x \gt 0#.

Using the chain rule, we get :

#f'(x) = 1/(2 sqrt(x)) sin(1/x) - 1/x^(3/2) cos(1/x) = (1/2 x sin(1/x) - cos(1/x))/x^(3/2)#

Using the identity #\abs(sin(y)) \leq y# for #y \geq 0#, we get (putting #y = 1/x#) :

#1/2 x sin(1/x) - cos(1/x) \leq 1/2 - cos(1/x)# which goes to #-1/2# when #x# goes to infinity.

In particular, when #x# is big enough, one has #f' \leq 0#, i.e. #f# is decreasing (so the sequence #u_n# is also decreasing for #n# big enough).

Now, let's calculate #lim_(n->+oo) u_n# :

#lim_(n->+oo) u_n = lim_(n->+oo) sqrt(n) * sin(1/n)#

You can't calculate the limit if #u_n# stay like this, it would give you #''(+oo)''*0#, which is undefined.

#lim_(n->+oo) sqrt(n) * sin(1/n) = lim_(n->+oo) (sin(1/n))/(1/sqrt(n))#

#=lim_(n->+oo) ((sin(1/n))')/((1/sqrt(n))')# (L'Hospital's Rule)

#=lim_(n->+oo) (cos(1/n)*(-1/n^2))/(-1/(2sqrt(n^3)))#

#=lim_(n->+oo) cos(1/n)*(1/n^2)*2sqrt(n^3)#

#=lim_(n->+oo) (2sqrt(n^3))/n^2 = lim_(n->+oo)(sqrt(n^3))/n^2#

#=lim_(n->+oo) x^(-1/2)=lim_(n->+oo)1/sqrt(x)= 0#.

Therefore, the alternating series #sum_(n=1)^(+oo)(-1)^n*u_n# converges conditionally.

Let's see if the series converges absolutely. It converges absolutely if the series of sequence #u_n# converges.

We will use the comparison test for #n>=1#.

#u_n = sqrt(n)*sin(1/n) >= sin(1/n) = v_n#

We know that the series of sequence #sin(1/n)# diverges because, by comparison test,

#lim_(n->+oo)v_n/h_n = lim_(n->+oo)sin(1/n)/(1/n)# (We use the harmonical series, which is divergent)

#=lim_(n->+oo) (sin(1/n)')/((1/n)')# (L'Hospital's Rule)

#=lim_(n->+oo) (cos(1/n)*(-1/n^2))/(-1/(n^2)) = lim_(n->+oo) cos(1/n) = 1#.

Since #lim_(n->+oo)v_n/h_n != 0# or #lim_(n->+oo)v_n/h_n != +oo#,

the series of sequence #v_n# and the series of sequence #h_n# are both convergent or divergent (by comparison test).
Since the series of sequence #h_n# is divergent, the series of sequence #v_n = sin(1/n)# is divergent too.

Since #u_n >= v_n# and the series of sequence #v_n# is divergent, the series of sequence #u_n = sqrt(n) * sin(1/n)# is divergent too (by comparison test).

Therefore, the alternating series #sum_(n=1)^(+oo)(-1)^n * u_n# doesn't converge absolutely.

Apr 20, 2017

Answer:

The series is convergent.

Explanation:

We know that

#(-1)^n sqrt(n)sin(1/n)=(-1)^nsqrtn (1/n)(sin(1/n)/(1/n)) = (-1)^n1/sqrtn (sin(1/n)/(1/n))#

so as #n# grows

#(-1)^n sqrt(n)sin(1/n) approx (-1)^n/sqrtn# and

#sum_n (-1)^n/sqrtn# converges because is an alternating series and
#abs(a_n)= 1/sqrtn# decreases monotonically to #0#

Finalizing, #sum_n(-1)^n sqrt(n)sin(1/n)# converges.