General Term of a series pi(3/11)^n+1?

Question

The general term of a series i tn=pi(3/11)^n+1. How many terms must be added to reach a sum that is greater than 0.32? The pi is the pi sign that equals 3.14....
This question is driving me nuts. the first term I get is larger than 0.32.

Answer 1

I'm assuming you mean π (3/11)^(n+1) since if the 1 is hanging off the end obviously all the terms are larger than 1 and you're done after one term. Also, I assume you're starting from n = 1 since if you start from n = 0 the first term is 3π/11 ≈ 0.85.
I get the following terms and partial sums:
0.23367 . . 0.23367
0.06373 . . 0.29740
0.01738 . . 0.31478
0.00474 . . 0.31952
0.00129 . . 0.32081
so five terms are required.

Note that you can solve this without evaluating each term, by noting that it's a geometric series with first term 9π/121 and common ratio 3/11. So the sum after n terms is
9π/121 (1 - (3/11)^n) / (1 - 3/11)
= (9π/121) (1 - (3/11)^n) (11/8)
= (9π/88) (1 - (3/11)^n).
For this to be equal to 0.32 we need
1 - (3/11)^n = 0.32 (88 / 9π) = 0.99596 to 5 d.p.
so (3/11)^n = 0.00404
and n log (3/11) = log 0.00404
=> n = log 0.00404 / log (3/11) = 4.24
So we need 5 terms to pass the target.