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.

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?

step-by-step explanation:

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