Decide whether each of the following series converges. if a given series converges, compute its sum. otherwise, enter inf if it diverges to infinity, minf if it diverges to minus infinity, and div otherwise.

1. ∑n=1[infinity](sin(10n)−sin(10n+1))
2. ∑n=1[infinity](e11n−e11(n+1))
3. ∑n=1[infinity](sin(10n)−sin(10(n+

1st answer is DIV 2nd answer is MNIF and 3rd answer is DIV

Step-by-step explanation:

1. ∑n=1[infinity](sin(10n)−sin(10n+1))

By applying formula sin C - sin D =2sin(C-D)/2cos(C+D)/2∑n=1[infinity](2sin(-1/2)cos(10n+1/2)By expanding the summation we get -2sin 1/2[ cos(10+1/2)+cos(20+1/2)+cos(30+1/2)+cos(40+1/2)+cos(50+1/2)+cos(60+1/2)+cos(70+1/2)+cos(80+1/2)-sin(1/2)-sin(10+1/2)-sin(20+1/2)-sin(30+1/2)-sin(40+1/2)-sin(50+1/2)-sin(60+1/2)-sin(70+1/2)-sin(80+1/2)-cos(1/2)-cos(10+1/2)-cos(20+1/2)sin(∞ )]All the term cancels out and sin(∞) and other last term is left out which value we don't know so the answer is DIV

2. ∑n=1[infinity](e∧11n−e∧11(n+1))

By expanding the summation we get= e∧11-e∧22+e∧22-e∧33+e∧33-e∧44-e∞=e∧11-e∞e∞>>>>>>>e∧11=-∞So the answer is MNIF

3. ∑n=1[infinity](sin(10n)−sin(10(n+1)))

By expanding the summation we get=sin(10)-sin(20)+sin(20)-sin(30)+sin(30)-sin(40)-sin(∞)=sin(10)-sin(∞)sin(∞) value we don't know so we can't decide the answer so the answer DIV

It’s going to be d. 4x^2-4x-19’