Determine whether the series is convergent or divergent by expressing sn as a telescoping sum (as in this example). [infinity] 6 n(n + 3) n = 1 a) convergent

b) divergent

If it is convergent, find its sum. (If the quantity diverges, enter DIVERGES.)

I assume the series is supposed to be

The summand can be expanded into partial fractions:

Then the sum is equivalent to

Consider the N-th partial sum of the series:

2[(1 - 1/4) + (1/2 - 1/5) + (1/3 - 1/6) + (1/4 - 1/7) + (1/5 - 1/8) + …

+ (1/(N - 3) - 1/N) + (1/(N - 2) - 1/(N + 1)) + (1/(N - 1) - 1/(N + 2)) + (1/N - 1/(N + 3))]

Each of the negative terms (in bold) will have a positive counterpart (underlined) that cancel, so the N-th partial sum would be

As N approaches infinity, the last three terms in the sum converge to 0, so the original sum converges to

b.an increase in a causes b to decrease