Geometric Series - Calculus 2

1. Evaluate the sum $\sum_{n=2}^{\infty} \frac{5 \cdot 3^{n-2}}{4^{n-1}}$.
   Answer: 5 First, rewrite the general term of the series to fit the form $ar^n$. $\frac{5 \cdot 3^{n-2}}{4^{n-1}} = \frac{5 \cdot 3^n \cdot 3^{-2}}{4^n \cdot 4^{-1}} = \frac{5 \cdot (1/9)}{(1/4)} (\frac{3}{4})^n = \frac{5}{9} \cdot 4 \cdot (\frac{3}{4})^n = \frac{20}{9} (\frac{3}{4})^n$. The series is $\frac{20}{9} \sum_{n=2}^{\infty} (\frac{3}{4})^n$. This is a geometric series with ratio $r = 3/4$. The sum starts at $n=2$, so the first term of $\sum_{n=2}^{\infty} (\frac{3}{4})^n$ is $(3/4)^2 = 9/16$. The sum of this part is $S = \frac{a}{1-r} = \frac{9/16}{1 - 3/4} = \frac{9/16}{1/4} = \frac{9}{4}$. The total sum is $\frac{20}{9} \cdot S = \frac{20}{9} \cdot \frac{9}{4} = 5$.
2. Which of the following describes the set of all $c$ for which the series $\sum_{n=1}^{\infty} (\frac{1}{1+c^2})^n$ converges?
   Answer: $c \ne 0$ This is a geometric series with ratio $r = \frac{1}{1+c^2}$. For the series to converge, we need $|r| < 1$. So, we must satisfy $|\frac{1}{1+c^2}| < 1$. Since $c^2 \ge 0$, the denominator $1+c^2$ is always greater than or equal to 1. This means $0 < \frac{1}{1+c^2} \le 1$. The convergence condition $|r|<1$ requires a strict inequality. Therefore, we need $\frac{1}{1+c^2} < 1$. This is true as long as the denominator $1+c^2 > 1$, which means $c^2 > 0$. This condition holds for all real numbers $c$ except for $c=0$. If $c=0$, the ratio $r=1$, and the series diverges. Thus, the series converges for all $c \ne 0$.
3. Let $a_n = \int_0^2 (x/3)^n dx$. Evaluate $\sum_{n=0}^{\infty} a_n$.
   Answer: $3\ln(3)$ We can swap the summation and integration because the series converges uniformly on the interval of integration. $\sum_{n=0}^{\infty} a_n = \sum_{n=0}^{\infty} \int_0^2 (x/3)^n dx = \int_0^2 \sum_{n=0}^{\infty} (x/3)^n dx$. The expression inside the integral is a geometric series with ratio $r = x/3$. For $x \in [0, 2]$, $|r| \le 2/3 < 1$, so the series converges to $\frac{1}{1 - x/3} = \frac{3}{3-x}$. Now, we evaluate the integral: $\int_0^2 \frac{3}{3-x} dx$. Using a u-substitution with $u=3-x$ and $du=-dx$, the integral becomes $\int_3^1 \frac{3}{-du} = -3 \int_3^1 \frac{1}{u} du = 3 \int_1^3 \frac{1}{u} du = 3[\ln|u|]_1^3 = 3(\ln(3) - \ln(1)) = 3\ln(3)$.
4. A geometric series is defined by $\sum_{n=2}^{\infty} \frac{3 \cdot 2^{n+1}}{5^n}$. What is the sum of this series?
   Answer: $\frac{8}{5}$ The terms of the series can be rewritten to identify the common ratio. $\frac{3 \cdot 2^{n+1}}{5^n} = \frac{3 \cdot 2 \cdot 2^n}{5^n} = 6(\frac{2}{5})^n$. This is a geometric series with common ratio $r = 2/5$. Since $|r| < 1$, the series converges. The summation starts at $n=2$, so the first term $a$ is $6(\frac{2}{5})^2 = 6(\frac{4}{25}) = \frac{24}{25}$. The sum of an infinite geometric series is $S = \frac{a}{1-r}$. Plugging in the values, $S = \frac{24/25}{1 - 2/5} = \frac{24/25}{3/5} = \frac{24}{25} \cdot \frac{5}{3} = \frac{8}{5}$.
5. A ball is dropped from a height of 10 meters. Each time it bounces, it reaches a height that is $3/4$ of the previous height. What is the total vertical distance traveled by the ball before it comes to rest?
   Answer: 70 m The total distance is the initial drop plus the sum of the distances of all the bounces (up and down). Initial drop: 10 m. First bounce (up and down): $10(\frac{3}{4}) + 10(\frac{3}{4}) = 20(\frac{3}{4})$. Second bounce: $10(\frac{3}{4})^2 + 10(\frac{3}{4})^2 = 20(\frac{3}{4})^2$. The total distance of the bounces is an infinite geometric series: $\sum_{n=1}^{\infty} 20(\frac{3}{4})^n$. The first term is $a = 20(3/4) = 15$ and the ratio is $r = 3/4$. The sum of the bounces is $S_{bounces} = \frac{a}{1-r} = \frac{15}{1 - 3/4} = \frac{15}{1/4} = 60$ m. The total vertical distance is the initial drop plus the sum of the bounce distances: $10 + 60 = 70$ m.
6. Let $f(x) = \frac{2x}{3}$. Define a sequence $x_n$ by $x_0=9$ and $x_{n+1} = f(x_n)$ for $n \ge 0$. Find the value of $\sum_{n=0}^{\infty} x_n$.
   Answer: 27 We can find the first few terms of the sequence to identify the pattern. $x_0 = 9$. $x_1 = f(x_0) = f(9) = \frac{2(9)}{3} = 6$. $x_2 = f(x_1) = f(6) = \frac{2(6)}{3} = 4$. The sequence is $9, 6, 4, \ldots$. This is a geometric sequence with first term $a = 9$ and common ratio $r = \frac{6}{9} = \frac{2}{3}$. The sum of the corresponding infinite series is $S = \frac{a}{1-r} = \frac{9}{1 - 2/3} = \frac{9}{1/3} = 27$.
7. A quantity $y$ changes over discrete time steps $n=0, 1, 2, \ldots$ such that $y_{n+1} = y_n - \frac{1}{4}y_n$, with $y_0=100$. What is the total sum of all values of $y$, $\sum_{n=0}^{\infty} y_n$?
   Answer: 400 The recurrence relation is $y_{n+1} = y_n - \frac{1}{4}y_n = (1 - \frac{1}{4})y_n = \frac{3}{4}y_n$. This shows that the sequence $\{y_n\}$ is a geometric sequence. The first term is $a = y_0 = 100$, and the common ratio is $r = 3/4$. We need to find the sum of the corresponding infinite series, $\sum_{n=0}^{\infty} 100(\frac{3}{4})^n$. Since $|r| = 3/4 < 1$, the series converges. The sum is $S = \frac{a}{1-r} = \frac{100}{1 - 3/4} = \frac{100}{1/4} = 400$.
8. A square has a side length of 4. A second square is inscribed by connecting the midpoints of the sides of the first square. A third square is inscribed in the second square in the same way, and this process continues indefinitely. What is the sum of the areas of all the squares?
   Answer: 32 The first square has area $A_1 = 4^2 = 16$. When a new square is inscribed by connecting the midpoints, its side length $s_2$ can be found using the Pythagorean theorem. A corner triangle has legs of length $4/2 = 2$. So, $s_2^2 = 2^2 + 2^2 = 8$. The area of the second square is $A_2 = s_2^2 = 8$. The ratio of the areas is $A_2/A_1 = 8/16 = 1/2$. This ratio holds for all subsequent squares. The sum of the areas is an infinite geometric series: $16 + 8 + 4 + \cdots$. The first term is $a=16$ and the common ratio is $r=1/2$. The sum is $S = \frac{a}{1-r} = \frac{16}{1-1/2} = \frac{16}{1/2} = 32$.
9. What is the sum of the series $\sum_{n=1}^{\infty} (\cos(\pi/3))^{2n}$?
   Answer: $\frac{1}{3}$ First, evaluate the constant term: $\cos(\pi/3) = 1/2$. The series becomes $\sum_{n=1}^{\infty} (1/2)^{2n}$. This can be rewritten as $\sum_{n=1}^{\infty} ((1/2)^2)^n = \sum_{n=1}^{\infty} (1/4)^n$. This is a geometric series with a common ratio $r = 1/4$. The summation starts at $n=1$, so the first term is $a = 1/4$. The sum is given by $S = \frac{a}{1-r} = \frac{1/4}{1 - 1/4} = \frac{1/4}{3/4} = \frac{1}{3}$.
10. What is the sum of the series $\sum_{n=1}^{\infty} (\frac{1}{e^n} - \frac{1}{e^{n+1}})$?
    Answer: $\frac{1}{e}$ This is a telescoping series. The $k$-th partial sum is $S_k = (\frac{1}{e} - \frac{1}{e^2}) + (\frac{1}{e^2} - \frac{1}{e^3}) + \cdots + (\frac{1}{e^k} - \frac{1}{e^{k+1}}) = \frac{1}{e} - \frac{1}{e^{k+1}}$. As $k \to \infty$, $\frac{1}{e^{k+1}} \to 0$, so the sum is $\frac{1}{e}$. Alternatively, splitting into geometric series: $\sum_{n=1}^{\infty} (\frac{1}{e})^n - \sum_{n=1}^{\infty} (\frac{1}{e})^{n+1} = \frac{1/e}{1-1/e} - \frac{1/e^2}{1-1/e} = \frac{1}{e-1} - \frac{1}{e(e-1)} = \frac{e-1}{e(e-1)} = \frac{1}{e}$.
11. Let $S_1 = \sum_{n=0}^{\infty} (x-2)^n$ and $S_2 = \sum_{n=0}^{\infty} (\frac{x}{3})^n$. For which open interval of $x$ values do both series converge?
    Answer: $(1, 3)$ For a geometric series to converge, the absolute value of its common ratio must be less than 1. For $S_1$, the ratio is $r_1 = x-2$. Convergence requires $|x-2| < 1$, which is equivalent to $-1 < x-2 < 1$. Adding 2 to all parts gives $1 < x < 3$. For $S_2$, the ratio is $r_2 = x/3$. Convergence requires $|x/3| < 1$, which is equivalent to $-1 < x/3 < 1$. Multiplying by 3 gives $-3 < x < 3$. For both series to converge, $x$ must be in the intersection of these two intervals: $(1, 3) \cap (-3, 3)$. The intersection is the interval $(1, 3)$.
12. Let $A = \sum_{n=1}^{\infty} (\frac{1}{x})^n$ and $B = \sum_{n=0}^{\infty} (\frac{1}{x+1})^n$. If $A=B$ and both series converge, what is the value of $x$?
    Answer: $\frac{1 + \sqrt{5}}{2}$ First, find the sums of the series in terms of $x$. For series A, $a = 1/x$ and $r = 1/x$. The sum is $A = \frac{1/x}{1 - 1/x} = \frac{1}{x-1}$. For A to converge, $|1/x| < 1$, which means $|x| > 1$. For series B, $a = 1$ and $r = 1/(x+1)$. The sum is $B = \frac{1}{1 - 1/(x+1)} = \frac{x+1}{x}$. For B to converge, $|1/(x+1)| < 1$, which means $|x+1| > 1$. This implies $x > 0$ or $x < -2$. The combined condition for convergence is $x > 1$ or $x < -2$. Now, set $A=B$: $\frac{1}{x-1} = \frac{x+1}{x}$. Cross-multiplying gives $x = (x-1)(x+1) = x^2 - 1$. This leads to the quadratic equation $x^2 - x - 1 = 0$. Using the quadratic formula, $x = \frac{1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{1 \pm \sqrt{5}}{2}$. We must check which solution satisfies the convergence condition. $x_1 = \frac{1 + \sqrt{5}}{2} \approx 1.618$, which is greater than 1, so it is a valid solution. $x_2 = \frac{1 - \sqrt{5}}{2} \approx -0.618$, which is not in the convergence interval. Thus, the only solution is $x = \frac{1 + \sqrt{5}}{2}$.
13. Let $S = \sum_{n=0}^{\infty} (\frac{-1}{2})^n$ and $T = \sum_{n=2}^{\infty} (\frac{1}{3})^n$. Find the value of $S \cdot T$.
    Answer: $\frac{1}{9}$ First, calculate the sum S. This is a geometric series with first term $a_S = (-1/2)^0 = 1$ and ratio $r_S = -1/2$. The sum is $S = \frac{1}{1 - (-1/2)} = \frac{1}{3/2} = \frac{2}{3}$. Next, calculate the sum T. This is a geometric series with ratio $r_T = 1/3$ that starts at $n=2$. The first term is $a_T = (1/3)^2 = 1/9$. The sum is $T = \frac{1/9}{1 - 1/3} = \frac{1/9}{2/3} = \frac{1}{9} \cdot \frac{3}{2} = \frac{1}{6}$. Finally, compute the product: $S \cdot T = \frac{2}{3} \cdot \frac{1}{6} = \frac{2}{18} = \frac{1}{9}$.
14. An infinite geometric series has a first term of 12 and a sum of 16. What is the sum of the first 3 terms of this series?
    Answer: $\frac{63}{4}$ First, find the common ratio $r$ using the formula for the sum of an infinite geometric series, $S = \frac{a}{1-r}$. We have $16 = \frac{12}{1-r}$. Solving for $r$ gives $16(1-r) = 12 \Rightarrow 1-r = \frac{12}{16} = \frac{3}{4} \Rightarrow r = \frac{1}{4}$. The sum of the first $n$ terms is given by $S_n = \frac{a(1-r^n)}{1-r}$. For the first 3 terms, $S_3 = \frac{12(1 - (1/4)^3)}{1 - 1/4} = \frac{12(1 - 1/64)}{3/4} = \frac{12(63/64)}{3/4} = 16 \cdot \frac{63}{64} = \frac{63}{4}$. Alternatively, the first three terms are $12$, $12(\frac{1}{4})=3$, and $3(\frac{1}{4})=\frac{3}{4}$. Their sum is $12 + 3 + \frac{3}{4} = 15 + \frac{3}{4} = \frac{63}{4}$.
15. Consider the geometric series $\sum_{n=0}^{\infty} \frac{2^{n+1}}{3^{n-1}}$. Which of the following correctly describes this series?
    Answer: Converges to 18 with first term 6 and common ratio $\frac{2}{3}$ When you encounter a geometric series, your first step is to rewrite it in standard form $\sum_{n=0}^{\infty} ar^n$ to identify the first term $a$ and common ratio $r$. Let's simplify the given series $\sum_{n=0}^{\infty} \frac{2^{n+1}}{3^{n-1}}$. Using exponent rules: $\frac{2^{n+1}}{3^{n-1}} = 2^{n+1} \cdot 3^{-(n-1)} = 2^{n+1} \cdot 3^{1-n} = 2^{n+1} \cdot 3^1 \cdot 3^{-n} = 3 \cdot 2^{n+1} \cdot 3^{-n} = 3 \cdot 2 \cdot 2^n \cdot 3^{-n} = 6 \cdot \left(\frac{2}{3}\right)^n$ So we have $\sum_{n=0}^{\infty} 6\left(\frac{2}{3}\right)^n$, which is geometric with first term $a = 6$ and common ratio $r = \frac{2}{3}$. Since $|r| = \frac{2}{3} < 1$, the series converges to $\frac{a}{1-r} = \frac{6}{1-\frac{2}{3}} = \frac{6}{\frac{1}{3}} = 18$. This confirms answer B. Looking at the wrong choices: A incorrectly claims the series diverges and misunderstands that a ratio less than 1 means terms decrease, not increase. C has the wrong first term (4 instead of 6) and wrong ratio ($\frac{3}{2}$ instead of $\frac{2}{3}$). D correctly identifies the first term as 6 but wrongly concludes divergence—the size of the first term doesn't determine convergence; only the ratio matters. Study tip: Always rewrite geometric series in standard form first. A series $\sum ar^n$ converges when $|r| < 1$, regardless of how large the first term is.
16. A geometric series $\sum_{k=2}^{\infty} ar^{k-1}$ converges to 20. If the series $\sum_{k=2}^{\infty} ar^{k+1}$ converges, what is its sum?
    Answer: $20r^2$, by multiplying each term by $r^2$ When you encounter geometric series problems with different starting indices or exponent modifications, focus on the relationship between the original and modified series rather than recalculating from scratch. The first series $\sum_{k=2}^{\infty} ar^{k-1}$ converges to 20. To find the sum of $\sum_{k=2}^{\infty} ar^{k+1}$, notice that each term in the second series can be written as $ar^{k+1} = ar^{k-1} \cdot r^2$. This means every term in the second series equals the corresponding term in the first series multiplied by $r^2$. Since multiplication by a constant distributes over infinite sums (when convergent), the second series converges to $20r^2$. This makes choice B correct. Let's examine why the other options fail. Choice A applies $\frac{20r^2}{1-r^2}$, which incorrectly treats this as a completely new geometric series and misapplies the standard formula $\frac{a}{1-r}$. The formula doesn't apply directly here since we're modifying an existing convergent series. Choice C gives the right numerical answer ($20r^2$) but provides weak reasoning—while the ratio between corresponding terms is constant ($r^2$), the explanation should focus on how constants factor out of convergent series. Choice D uses $\frac{20}{r^2}$, which would correspond to dividing each term by $r^2$ instead of multiplying. Study tip: When modifying convergent series by multiplying each term by a constant, the new sum equals the original sum times that constant. This property lets you solve these problems quickly without recalculating the entire series formula.
17. The series $\sum_{n=1}^{\infty} \frac{2}{(k+1)^n}$ converges to $1/3$. What is the value of $k$?
    Answer: 6 The given series is $2 \sum_{n=1}^{\infty} (\frac{1}{k+1})^n$. This is a geometric series with common ratio $r = \frac{1}{k+1}$. Since the series starts at $n=1$, the first term is $a = \frac{1}{k+1}$. The sum of this part of the series is $S = \frac{a}{1-r} = \frac{1/(k+1)}{1 - 1/(k+1)} = \frac{1/(k+1)}{(k+1-1)/(k+1)} = \frac{1}{k}$. The total sum of the original series is $2S = \frac{2}{k}$. We are given that this sum is $1/3$. Therefore, $\frac{2}{k} = \frac{1}{3}$, which implies $k = 6$. For convergence, we need $|r|<1$, so $|\frac{1}{k+1}|<1$, which means $k>0$ or $k<-2$. Our solution $k=6$ satisfies this condition.
18. The sum of the infinite series $\sum_{n=1}^{\infty} \frac{5}{(-3)^n}$ is approximated by its fourth partial sum, $S_4$. What is the value of the remainder, $R_4 = S - S_4$?
    Answer: $-\frac{5}{324}$ The series is a geometric series with first term (at n=1) $a = \frac{5}{-3} = -\frac{5}{3}$ and common ratio $r = -\frac{1}{3}$. The remainder $R_4$ is the sum of all terms after the 4th term, which is the series $\sum_{n=5}^{\infty} \frac{5}{(-3)^n}$. The first term of this new series is $a_5 = \frac{5}{(-3)^5} = -\frac{5}{243}$. The ratio is still $r = -\frac{1}{3}$. The sum of this remainder series is $R_4 = \frac{\text{first term}}{1-r} = \frac{-5/243}{1 - (-1/3)} = \frac{-5/243}{4/3} = -\frac{5}{243} \cdot \frac{3}{4} = -\frac{5}{81 \cdot 4} = -\frac{5}{324}$.
19. Consider the infinite geometric series $\sum_{k=0}^{\infty} \frac{(-1)^k \cdot 3^{k+1}}{4^k}$. Which statement about this series is correct?
    Answer: The series converges to $\frac{12}{7}$ because the common ratio is $-\frac{3}{4}$ Rewrite the series as $\sum_{k=0}^{\infty} 3 \cdot \frac{(-1)^k \cdot 3^k}{4^k} = 3\sum_{k=0}^{\infty} \left(-\frac{3}{4}\right)^k$. This is a geometric series with first term $a = 3$ and common ratio $r = -\frac{3}{4}$. Since $|r| = \frac{3}{4} < 1$, the series converges to $\frac{a}{1-r} = \frac{3}{1-(-\frac{3}{4})} = \frac{3}{1+\frac{3}{4}} = \frac{3}{\frac{7}{4}} = \frac{12}{7}$. Choice B is wrong because $|r| < 1$. Choice C misidentifies the first term. Choice D incorrectly identifies both the first term and ratio.
20. Two people, A and B, take turns flipping a fair coin, with A going first. The first person to flip a head wins. What is the probability that person B wins?
    Answer: $\frac{1}{3}$ Person B can win on their first turn (the 2nd flip), their second turn (4th flip), and so on. For B to win on the 2nd flip, the sequence of outcomes must be Tails, Heads (TH). The probability is $(1/2)(1/2) = 1/4$. For B to win on the 4th flip, the sequence must be THTH, with probability $(1/2)^4 = 1/16$. For B to win on the 6th flip, the sequence must be THTHTH, with probability $(1/2)^6 = 1/64$. The total probability of B winning is the sum of these probabilities: $P(B \text{ wins}) = \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \cdots$. This is an infinite geometric series with first term $a = 1/4$ and common ratio $r = 1/4$. The sum is $S = \frac{a}{1-r} = \frac{1/4}{1 - 1/4} = \frac{1/4}{3/4} = \frac{1}{3}$.

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