Why is a convergent and b divergent?

Part (a)

This sequence is convergent because all we care about are the leading terms for the numerator and denominator. As n gets really large, the leading terms dictate what happens.

The leading term up top is 9n

The leading term down below is 7n

Divide the two items to get (9n)/(7n) = 9/7

We don't have to worry about dividing by zero because n will go off to infinity, and as it does so, the sequence converges to 9/7

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Part (b)

As n heads to infinity, the terms of the sequence are approaching 9/7 (they never quite hit the exact value though).

Note how 9/7 = 1.2857 approximately which is larger than 0

So effectively for very large values of n, we're adding on 1.2857 each time or close to it, meaning that this infinite series diverges.

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For a series like this to converge, the answer to part (a) must be 0. This way at some point, we're effectively adding 0 to the series and we approach some fixed sum.

For example, consider the sequence 1/(7n+1)

As n goes to infinity, the terms of the sequence approach 0

This means that for some very large n, the 1/(7n+1) is basically 0 and we aren't adding much onto the sum to change it. So this sum will converge to a fixed value.

answer: haha (: i’ll go see what it is

step-by-step explanation:

i believe the answer is letter c) 8

step-by-step explanation: