Answer:
The only positive integers would be anything less than 3. So 1 or 2 would work
Step-by-step explanation:
If something that is already being raised to a power, gets raised to another power, those powers add up. So if it were , the exponents would add which would make it
.
In the case of your question, if 'n' was 2 it would be to the power of 6 making it less than to the power of. 1 could also work
All three series converge, so the answer is D.
The common ratios for each sequence are (I) -1/9, (II) -1/10, and (III) -1/3.
Consider a geometric sequence with the first term <em>a</em> and common ratio |<em>r</em>| < 1. Then the <em>n</em>-th partial sum (the sum of the first <em>n</em> terms) of the sequence is
Multiply both sides by <em>r</em> :
Subtract the latter sum from the first, which eliminates all but the first and last terms:
Solve for :
Then as gets arbitrarily large, the term will converge to 0, leaving us with
So the given series converge to
(I) -243/(1 + 1/9) = -2187/10
(II) -1.1/(1 + 1/10) = -1
(III) 27/(1 + 1/3) = 18
