Question

Show that the series sin[(-1)^n/n] converges conditionally ?

Answer 1

i've already proved that lim of it approaches 0, but how can we decide it to be convergent ?

Answer 2

i mean, convergent then lim = 0 but it's not true vice versa

Answer 3

if the limit of the sequence approaches 0, then we can not conclude the series is convergent. One example is 1/n.

∑sin( (-1)^n / n) is a alternating series yes?

Answer 4

yes, it's alternating series, but only the (-1)^n/n is convergent. How can we combine with sine function?

Answer 5

well, use the fact that sin(-a) = -sin(a)

we have ∑sin( (-1)^n / n) =∑ (-1)^n sin(1/n) , yes?

Answer 6

yes, it approaches 0, but how to prove it's convergent ?

Answer 7

the criteria to test whether an alternating series is convergent is:
1) a(n+1) <= a(n)
2) lim a(n) = 0
n-> 0

Answer 8

you already showed the second part. Now you need to show that
sin(1/(n+1)) <= sin(1/n)

Answer 9

so we have
1/(n+1) <= 1/n (which is true)

since both criteria meets, by thy Alternating Series Test, the series converge.

Answer 10

to so the series converge conditionally, show that ∑ |a(n)| does not converge

Answer 11

okay, i see

Answer 12

| (-1)^n sin(1/n) | = sin(1/n)

compare with (1/n)

lim sin(1/n) / (1/n) = 1
n->inf

which means both series either converge or diverge. But 1/n diverges, hence, sin(1/n) also diverges

Answer 13

So the series converge conditionally

Answer 14

But how can sin(1/n) /(1/n) = 1 ?

Answer 15

L'hospital rule

Answer 16

I see, thank you!