Determine whether the series is convergent or divergent.
Σ[infinity] n=1 7 sin 1/n
a. The Limit Comparison Test allows us to determine convergence or divergence by considering lim n-->[infinity] an/bn.
We will use an = sin 1/n and bn = 1/n .
The terms 1/n are positive since n is positive. Since 0 < 1/n < π/2, then the terms sin 1/n are .

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

in simplified version   it would be x ≤-4

step-by-step explanation:

multiply each term in x ≤-4 by -1

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step-by-step explanation: