Answer:
Step-by-step explanation:
Test for convergence or divergence of a series using Comparison test
Function.
Σ Sin(1/n). From n= 1 to n=∞
Comparison test states that
Let, 0 ≤ an ≤ bn
For all n
Then, if
Σbn converges then, Σan converge
Σan diverges then, Σbn diverge
I.e if the small series diverges, then, the big series diverges and if the big series converges then, the small series converges.
Let compare the given function to 1/n
1/n is greater than Sin(1/n)
0 ≤ Sin(1/n) ≤ 1/n
1/n is a positive number and it is always greater than 1.
We know that 1/n diverse from p series
P series says that,
Σ 1/n^p
If p≤1 then the series diverges
If p>1 the series converges
So when we compare this to 1/n, then p is equal to 1, therefore, the series diverges.
Now, comparing this to the comparison thereom
Since, the large series bn diverges, then the smaller series an diverges.
Then,
Σ 1 / n diverges implies that.
Σ Sin(1/n) diverges too
, 
, and 
, so at this point we get