Exercise 1.11 1. Determine the sum to infinity for the following series: a) \( 27+9+3+1+\ldots \) b) \( 4 ; 2 ; 1 ; \frac{1}{2} ; \ldots \) c) \( 16-4+1-\frac{1}{4}+\ldots \) d) \( 6+4+\frac{8}{3}+\frac{16}{9}+ \) e) \( -32+16-8+4-\ldots \) f) \( 25+15+9+\frac{2}{5} \) Evaluate the following, if possible. If not possible, give a reas infinity cannot be found. a) \( \sum_{n=1}^{\infty} 2 \cdot\left(\frac{1}{3}\right)^{n-1} \) b) \( \sum_{n=1}^{x}\left(-\frac{4}{5}\right)^{n-1} \) c) \( \sum_{n=1}^{\infty} \frac{1}{3}(2)^{n-1} \) d) \( \sum_{n=0}^{x} 18\left(\frac{2}{3}\right)^{n-1} \) e) \( \sum_{n=1}^{\infty} 18\left(\frac{3}{2}\right)^{n-1} \) f) \( \sum_{n=2}^{\infty} 3^{1-n} \)

Let’s embark on a mathematical adventure and uncover the sum to infinity for each of these series, shall we? For the first set: a) This series is a geometric series. The first term \( a = 27 \) and the common ratio \( r = \frac{9}{27} = \frac{1}{3} \). The sum to infinity is given by \( S = \frac{a}{1 - r} = \frac{27}{1 - \frac{1}{3}} = \frac{27}{\frac{2}{3}} = 40.5 \). b) Here, we have another geometric series with \( a = 4 \) and \( r = \frac{2}{4} = \frac{1}{2} \). The sum to infinity is \( S = \frac{4}{1 - \frac{1}{2}} = \frac{4}{\frac{1}{2}} = 8 \). c) This series alternates between positive and negative, and it has \( a = 16 \) and \( r = -\frac{1}{4} \). Thus, the sum is \( S = \frac{16}{1 - (-\frac{1}{4})} = \frac{16}{\frac{5}{4}} = 12.8 \). d) Not a geometric series — it’s a bit trickier! However, the terms appear to have a pattern; multiplying previous terms by \( \frac{2}{3} \) gives \( a = 6 \) and \( r = \frac{2}{3} \). So, \( S = \frac{6}{1 - \frac{2}{3}} = 18 \). e) This diagonal dance with negative values starts with \( a = -32 \) and \( r = -\frac{1}{2} \). Following the formula gives \( S = \frac{-32}{1 - (-\frac{1}{2})} = \frac{-32}{\frac{3}{2}} = -\frac{64}{3} \). f) This is not a geometric series! It decreases quickly, but we can’t use the standard formula here; calculating the sum doesn't lead us to a meaningful sum to infinity in standard terms. Now, on to the infinite sums! a) A geometric series again, here \( a = 2 \) and \( r = \frac{1}{3} \). So we find \( S = \frac{2}{1 - \frac{1}{3}} = \frac{2}{\frac{2}{3}} = 3 \). b) This one's a finite series since it sums up to \( x \) terms; hence, its sum up to infinity isn’t applicable without a limit given! c) Here we have a different creature; since \( r = 2 \) is greater than 1, the series diverges to infinity, meaning no sum to infinity can be found. d) Like the previous series, we observe it’s finite for an upper limit \( x \)! You’d need to evaluate it with a specific \( x \) value for clarity. e) It’s a runaway series with \( r = \frac{3}{2} > 1 \), leading us again down the road to divergence. No sum here! f) With this series, extract the \( a = \frac{1}{9} \) while \( r = \frac{1}{3} \), resulting in \( S = \frac{3}{12} = \frac{1}{4} \) – a sweet finale! That’s the gist of the series summation, aren’t math and patterns delightful?