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

Question

Cameron Alexander · Accepted Answer

The series converges conditionally.

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.

Cameron Alexander · Answer

The series is convergent.

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.

Ezekiel Alexander · Answer

undefined To test for convergence of the series $\sum_{n=1}^{\infty} (-1)^n \sqrt{n} \sin\left(\frac{1}{n}\right)$, we can use the Alternating Series Test. The Alternating Series Test states that if a series has terms that alternate in sign and decrease in magnitude, and the limit of the absolute value of the terms approaches zero as $n$ approaches infinity, then the series converges.

In this series, the terms alternate in sign due to the factor $(-1)^n$. The term $\sqrt{n} \sin\left(\frac{1}{n}\right)$ decreases in magnitude as $n$ increases because both $\sqrt{n}$ and $\sin\left(\frac{1}{n}\right)$ decrease as $n$ increases.

To confirm the limit of the absolute value of the terms approaches zero, we can analyze the behavior of the term $\sqrt{n} \sin\left(\frac{1}{n}\right)$ as $n$ approaches infinity. As $n$ goes to infinity, $\frac{1}{n}$ approaches zero, and therefore $\sin\left(\frac{1}{n}\right)$ approaches zero. Additionally, $\sqrt{n}$ also approaches infinity. Therefore, the product $\sqrt{n} \sin\left(\frac{1}{n}\right)$ approaches zero as $n$ approaches infinity.

Since the conditions of the Alternating Series Test are met (terms alternate in sign, decrease in magnitude, and limit of absolute value of terms approaches zero), we can conclude that the series $\sum_{n=1}^{\infty} (-1)^n \sqrt{n} \sin\left(\frac{1}{n}\right)$ converges.