Tom K. answered 03/27/21
Knowledgeable and Friendly Math and Statistics Tutor
To show that it is convergent by the alternating series test, we note that the sign of the sequence of terms alternates and 3/n^2 converges monotonically to 0.
With such series, the error is less than the absolute value of the next term.
Thus, for an error less than .05, we set 3/n^2 = .05
n^2 = 60
n = 7.75
As the absolute value of the term for n = 8 is less than .05, n = 7
It turns out that using the next term overestimates the number of terms required. 5 is sufficient.