- #1

- 590

- 0

i have a question where i need to solve the following series,

[tex]\sum[/tex](-1)

^{n-1}[tex]\frac{1}{n - sin(n)}[/tex] from 1 to infinity

i take the absolute value of An, and get An=[tex]\frac{1}{n-sin(n)}[/tex] now i take Bn=1/n which i know diverges with the limits 1, infinity.

lim(n->inf) [tex]\frac{Bn}{An}[/tex]=[tex]\frac{n-sin(n)}{n}[/tex]=1 therefore An behaves like Bn ==> diverges

if i take the abs values of the whole series i get i series which diverges leaving me with only 2 options, divergence or conditional convergance.

from here the only way i know of i leibniz laws,

-lim An =0

-An>A(n+1)

for the limAn =0 exists, but as for An>A(n+1) i cannot prove this since An is dependant on sin(n) which can be anywhere between -1 and 1 and is not necessarily larger or smaller for n+1 than for n. because of this i would think that the series diverges, but this is not the case. how do i prove that the series converges??