Chapter 10 Challenge Problems Solutions

SOLUTION Let's begin by supposing that \(\{a_n\}\) converges—that is, by assuming \(\lim_{n \to \infty} a_n\) \(= L\) for any real \(L \ne 0.\) We therefore have \(a_n \to L\) and \(a_{n + 1} \to L\) as \(n \to \infty.\) Taking the limits of both sides of the recursive relationship, we have \[ \ba \lim_{n \to \infty} a_{n + 1} &= \lim_{n \to \infty} \par{-\frac{1}{a_n}} \nl L &= -\frac{1}{L} \nl L^2 &= -1 \pd \ea \] But the equation \(L^2 = -1\) has no real solutions. Thus, we have a contradiction—no real value of \(L\) exists. So the sequence \(\{a_n\}\) cannot be convergent; it must instead diverge.

SOLUTION

1. If \(a_n = n\) and \(b_n = n,\) then both \[\sum_{n = 1}^\infty a_n \and \sum_{n = 1}^\infty b_n\] diverge by the Divergence Test because both \(\lim_{n \to \infty} a_n\) and \(\lim_{n \to \infty} b_n\) approach \(\infty.\) Then \[S = \sum_{n = 1}^\infty n + \sum_{n = 1}^\infty n \pd\] The terms of \(S\) are \[\par{1 + 2 + 3 + \cdots} + \par{1 + 2 + 3 + \cdots} \cma\] a divergent series. Thus, \(a_n = n\) and \(b_n = -n\) represent one satisfactory option.

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2. If \(a_n = n\) and \(b_n = -n,\) then both \[\sum_{n = 1}^\infty a_n \and \sum_{n = 1}^\infty b_n\] diverge by the Divergence Test because both \(\lim_{n \to \infty} a_n\) and \(\lim_{n \to \infty} b_n\) do not equal \(0.\) These choices of \(a_n\) and \(b_n\) show \[ \ba S &= \sum_{n = 1}^\infty n + \sum_{n = 1}^\infty (-n) \nl &= \par{1 + 2 + 3 + \cdots} - \par{1 + 2 + 3 + \cdots} \nl &= 0 \pd \ea \] So \(S\) is convergent. Thus, \(a_n = n\) and \(b_n = -n\) represent one satisfactory option.

SOLUTION To prove that the test is inconclusive, we simply need to find two contradictory examples, in each of which \(\sum a_n\) has a different conclusion when \(\sum b_n\) diverges. If \(a_n = 1/n^3\) and \(b_n = 1/n,\) then \(\sum a_n\) converges and \[L = \lim_{n \to \infty} \frac{a_n}{b_n} = 0 \pd \] But if \(a_n = 1/n\) and \(b_n = 1/\sqrt n,\) then \(\sum a_n\) and \(\sum b_n\) both diverge, and \[L = \lim_{n \to \infty} \frac{a_n}{b_n} = 0 \pd \] Thus, for \(L = 0\) the Limit Comparison Test fails to provide consistent conclusions about \(\sum a_n.\)

SOLUTION

1. The total money produced is the sum of lent money: \[ \ba M &= x(1 - R) + x(1 - R)^2 + x(1 - R)^3 + \cdots \nl &= \frac{x(1 - R)}{1 - (1 - R)} \nl &= \boxed{\frac{x(1 - R)}{R}} \ea \]

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2. The difference in the values of \(M\) is \[\frac{x(1 - (R + 0.1))}{R + 0.1} - \frac{x(1 - R)}{R} = \boxed{\frac{x(0.9 - R)}{R + 0.1} - \frac{x(1 - R)}{R}}\]

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3. The limit is \[\lim_{R \to 0^+} \frac{x(1 - R)}{R} \pd\] As \(R \to 0^+,\) the numerator approaches \(x\) and the denominator approaches \(0.\) The fraction therefore approaches positive infinity, so the limit equals \(\boxed{\infty}.\) Thus, if the reserve ratio is \(0,\) then infinite money is added to the money supply. In reality, however, loans are not made indefinitely; during times of recession, overall economic activity is slowed and so fewer bank transactions are seen.

SOLUTION

1. After \(1\) second, \(60\) frames have elapsed. Thus, the \(1\%\) reduction in the particle's speed is applied \(60\) times. The new velocity is therefore \[10(0.99)^{60} = \boxed{5.472 \undiv{m}{sec}}\]

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2. Note that each frame is \(1/60\) of a second. Also recall that distance is the product of speed and time. In the first frame, the particle travels a distance of \(10(1/60).\) In the second frame, the particle's new speed is \(10 \times 0.99,\) so it travels a distance of \((10 \times 0.99) (1/60).\) Likewise, in the third frame the particle's speed is \(10 \times 0.99^2,\) meaning its distance traveled is \((10 \times 0.99^2)(1/60).\) In general, in the \(n\)th frame (starting at \(n = 0\)) the particle's speed is \(10 \times 0.99^n,\) and its distance traveled is \((10 \times 0.99^n)(1/60).\) Hence, the total distance traveled is \[ \ba D_f &= (10) \par{\tfrac{1}{60}} + (10 \times 0.99) \par{\tfrac{1}{60}} + (10 \times 0.99^2) \par{\tfrac{1}{60}} + \cdots \nl &= \sum_{n = 0}^\infty (10 \times 0.99^n) \par{\tfrac{1}{60}} \nl &= \tfrac{1}{6} \sum_{n = 0}^\infty 0.99^n \nl &= \frac{\tfrac{1}{6}}{1 - 0.99} \nl &\approx \boxed{16.667 \un{m}} \ea \]

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3. Under the \(2\%\) condition, the particle disappears once its velocity reaches \(10 \times 0.02\) \(= 0.2 \undiv{m}{sec}.\) The particle's speed after the \(N\)th iteration is \(10 \times 0.99^N,\) so we see \[ \ba 10 \times 0.99^N &= 0.2 \nl \implies N &= \log_{0.99} 0.02 \approx 389.24 \pd \ea \] Since \(N\) is an integer, we round up to \(N = 390.\) (On the \(390\)th iteration, the particle's speed becomes less than \(2\%\) of its original.) Then the distance traveled over \(390\) frames—that is, from \(0\) to \(389\)—is \[ \baat{2} D_e &= \sum_{i = 0}^{389} (10 \times 0.99^i) \par{\tfrac{1}{60}} \nl &= \tfrac{1}{6} \sum_{i = 0}^{389} 0.99^i \nl &= \frac{\tfrac{1}{6} (1 - 0.99^{390})}{1 - 0.99} \nl &\approx \boxed{16.336 \un{m}} \eaat \]

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4. We see \[ \frac{D_e}{D_f} \times 100 \% \approx \frac{16.336}{16.667} \times 100\% = \boxed{98.014 \%} \] Thus, the developer's choice of making the particle disappear once it reaches \(2\%\) of its initial speed is appropriate; doing so allows the particle to travel nearly its full distance, \(D_f.\)

SOLUTION Because the second term of the series is \(2,\) the first term must be \(2/r,\) where \(r\) is the common ratio and \(0 \lt r \lt 1\) for convergence to a positive number. Thus, \[S = \frac{2/r}{1 - r} = \frac{2}{r - r^2} \pd\] Now we minimize \(S\) on \((0, 1).\) Taking its derivative and equating it to \(0,\) we attain \[\deriv{S}{r} = \frac{-2(1 - 2r)}{(r - r^2)^2} = 0 \implies r = \tfrac{1}{2} \pd\] This value corresponds to an absolute minimum of \(S,\) as you may verify. (See Section 3.1.) Hence, the smallest possible value is \[S_{\text{min}} = \frac{2}{\frac{1}{2} - \par{\frac{1}{2}}^2} = \frac{2}{1/4} = \boxed 8\]

SOLUTION Noting the pattern and regrouping the terms, we attain \[ \ba S &= \par{\tfrac{1}{4} + \tfrac{1}{16} + \tfrac{1}{64} + \cdots} + \par{\tfrac{2}{3} - \tfrac{2}{9} + \tfrac{2}{27} + \cdots} + \par{\tfrac{5}{6} - \tfrac{5}{12} + \tfrac{5}{24} + \cdots} \nl &= \sum_{n = 0}^\infty \tfrac{1}{4} \par{\tfrac{1}{4}}^n + \sum_{n = 0}^\infty \tfrac{2}{3} \par{-\tfrac{1}{3}}^n + \sum_{n = 0}^\infty \tfrac{5}{6} \par{-\tfrac{1}{2}}^n \pd \ea \] All three geometric series are convergent because their common ratios are each between \(-1\) and \(1.\) The sum of three convergent series is also convergent, so \[ \ba S &= \frac{1/4}{1 - \tfrac{1}{4}} + \frac{2/3}{1 - \par{-\tfrac{1}{3}}} + \frac{5/6}{1 - \par{-\tfrac{1}{2}}} \nl &= \tfrac{1}{3} + \tfrac{1}{2} + \tfrac{5}{9} \nl &= \boxed{\tfrac{25}{18}} \ea \]

SOLUTION

1. Using a right Riemann sum with \(\Delta x = 1,\) we approximate the area under \(y = 1/x\) from \(x = 1\) to \(x = N\) to be \[ \Delta x \par{\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots + \frac{1}{N}} = \sum_{i = 2}^N \frac{1}{i} \pd \] Since \(y = 1/x\) is decreasing, the Riemann sum is an underestimate of the area. We therefore have \[\sum_{i = 2}^N \frac{1}{i} \leq \int_1^N \frac{1}{x} \di x = \ln N \pd\] Adding \(1\) to both sides shows \[ \ba 1 + \sum_{i = 2}^N \frac{1}{i} &\leq 1 + \ln N \nl \sum_{i = 1}^N \frac{1}{i} &\leq 1 + \ln N \cma \ea \] as requested.

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2. The sum of the first million terms is given by \(N = 1\cmaNum000\cmaNum000.\) The inequality then gives \[\sum_{i = 1}^{1\cmaNum000\cmaNum000} \frac{1}{i} \leq 1 + \ln 1\cmaNum000\cmaNum000 \approx 14.816 \lt 15 \cma\] as requested. Likewise, the sum of the first billion terms is given by \(N = 1\cmaNum000\cmaNum000\cmaNum000 \col\) \[\sum_{i = 1}^{1\cmaNum000\cmaNum000\cmaNum000} \frac{1}{i} \leq 1 + \ln 1\cmaNum000\cmaNum000\cmaNum000 \approx 21.723 \lt 22 \cma\] as requested.

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3. Using the inequality from part (a), we get \[ \ba 50 \leq 1 + \ln N &\iffArrow 49 \leq \ln N \nl &\iffArrow N \geq e^{49} \approx \boxed{1.907 \times 10^{21}} \ea \] This is painfully large. The Harmonic series' slow divergence rate is explained by the gradual growth of the natural logarithm function.

SOLUTION First note that the \(p\)-series converges if and only if \(c^2 - 3 \gt 1\)—that is, \(c^2 \gt 4,\) meaning \(c \in (-\infty, -2) \cup (2, \infty).\) In addition, if \(c \notin [1, \infty),\) then the improper integral becomes \[-\frac{1}{x - c} \intEval_1^\infty = \frac{1}{1 - c} \cma\] which converges. If \(c \in [1, \infty),\) then the vertical asymptote at \(x = c\) would be included in the region of integration, and so the integral would diverge. Thus, the improper integral and infinite series both converge for \(c\) in \[\boxed{(-\infty, -2)}\]

SOLUTION The integral is equivalent to the following infinite series: \[ \ba \sum_{n = 1}^\infty \frac{(-1)^n}{(n - 1)!} \int_0^x t^{n - 1} \di t &= \sum_{n = 1}^\infty \frac{(-1)^n}{(n - 1)!} \frac{t^n}{n} \intEval_0^x \nl &= \sum_{n = 1}^\infty \frac{(-1)^n}{(n - 1)!} \frac{x^n}{n} \nl &= \sum_{n = 1}^\infty \frac{(-1)^n}{n!} \, x^n \cma \ea \] where the last step is true because \(n!\) \(= n(n - 1)!.\) The series can be rewritten as \[\sum_{n = 1}^\infty \frac{(-x)^n}{n!} \cma\] which resembles the Maclaurin series for \(e^{-x}.\) (See Section 10.8.) The only difference is the starting index; we have \[ \ba \sum_{n = 0}^\infty \frac{(-x)^n}{n!} &= e^{-x} \nl 1 + \sum_{n = 1}^\infty \frac{(-x)^n}{n!} &= e^{-x} \nl \implies \sum_{n = 1}^\infty \frac{(-x)^n}{n!} &= e^{-x} - 1 \cma \ea \] as requested.

SOLUTION

1. Let \(P(n)\) be a polynomial as a function of \(n.\) If the polynomial is nonconstant—that is, \(\deg P \gt 0\)—then \(\lim_{n \to \infty} P(n)\) must be either \(-\infty\) or \(\infty.\) Alternatively, if \(P\) is a nonzero constant \(K,\) then \(P \to K\) as \(n \to \infty.\) Because \(\lim_{n \to \infty} P(n) \ne 0,\) the Divergence Test asserts that \(\sum P(n)\) is divergent.
2. As \(n \to \infty,\) the ratio \(\abs{P(n + 1)/P(n)}\) is dominated by the highest-degree terms of the numerator and denominator: \(P(n + 1)\) and \(P(n),\) respectively. But both functions share the same degree and leading coefficient. Hence, \(\lim_{n \to \infty} \abs{P(n + 1)/P(n)} = 1\) and so the Ratio Test is inconclusive.
3. With the Ratio Test, we have \[ \ba L &= \lim_{n \to \infty} \abs{\frac{P(n + 1) \cdot e^{-n} \, e^{-1}}{P(n) \cdot e^{-n}}} \nl &= \lim_{n \to \infty} \abs{\frac{P(n + 1)}{P(n) \, e}} \pd \ea \] The ratio \(P(n + 1)/P(n)\) tends to \(1\) as \(n \to \infty,\) because the limit becomes the ratio of coefficients of the highest-degree terms. Hence, we have \(L = 1/e.\) Because \(L \lt 1,\) we conclude that \(\sum P(n) \, e^{-n}\) is absolutely convergent.

SOLUTION

1. Differentiating the power series gives \[f'(x) = \sum_{n = 1}^\infty \frac{n x^{n - 1}}{n!} = \sum_{n = 1}^\infty \frac{x^{n - 1}}{(n - 1)!} \pd\] Reindexing the series to begin at \(n = 0\)—in doing so, we replace the \((n - 1)\) terms with \(n\)—gives \[f'(x) = \sum_{n = 0}^\infty \frac{x^n}{n!} \pd\] But this power series is \(f(x).\) Hence, \(f'(x) = f(x).\)

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2. Let \(y = f(x).\) Then the differential equation of part (a) becomes \[\deriv{y}{x} = y \cma\] which we solve by Separation of Variables. We obtain \[ \ba \int \frac{\dd y}{y} &= \int \dd x \nl \ln \abs y &= x + C_1 \nl y &= Ce^x \nl f(x) &= Ce^x \cma \ea \] where \(C = \pm e^{C_1}.\) To solve for \(C,\) we note that \(f(0) = \sum_{n = 0}^\infty 0^n/n! = 1\) (assuming \(0^0 = 1\)). Hence, substituting the point \((0, 1)\) gives \[f(0) = Ce^0 = 1 \implies C = 1 \pd\] Thus, \(f(x) = e^x.\)

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3. We replace \(x\) with \(2x\) to get \[f(2x) = \boxed{\sum_{n = 0}^\infty \frac{(2x)^n}{n!}}\]

SOLUTION Note that \(\sum_{n = 1}^\infty a_n\) converges absolutely if and only if \(\sum_{n = 1}^\infty \abs{a_n}\) converges. Observe that \(\sum \abs{a_n}\) is telescoping, as shown by the following partial sum: \[ \ba \sum_{i = 1}^N \abs{a_i} &= \sum_{i = 1}^N \par{b_i - b_{i + 1}} \nl &= \par{b_1 - \cancel{b_2}} + \par{\cancel{b_2} - \cancel{b_3}} + \cdots + \par{\cancel{b_N} - b_{N + 1}} \nl &= b_1 - b_{N + 1} \pd \ea \] Hence, we find \[\sum_{n = 1}^\infty \abs{a_n} = \lim_{N \to \infty} \sum_{i = 1}^N \abs{a_i} = \lim_{N \to \infty} \par{b_1 - b_{N + 1}} = b_1 \cma\] where the last step is true because \(b_N\) (and therefore \(b_{N + 1}\)) is given to approach \(0\) as \(N \to \infty.\) Because \(\sum_{n = 1}^\infty \abs{a_n}\) converges, \(\sum_{n = 1}^\infty a_n\) converges absolutely, as requested.

SOLUTION Because \(n \ne 1,\) \[ \ba \int_1^\infty \frac{1}{x^n} \di x &= \int_1^\infty x^{-n} \di x \nl &= \frac{x^{1 - n}}{1 - n} \intEval_1^\infty \nl &= \frac{1}{n - 1} \cma \ea \] where the last step is true because \(x^{1 - n} \to 0\) as \(x \to \infty\) (since the power \(1 - n\) is negative.) So the given series becomes \[\sum_{n = 2}^\infty \frac{1}{n - 1} \pd\] Shifting the index, we have \[\sum_{n = 1}^\infty \frac{1}{n} \cma\] the divergent Harmonic series. Thus, sum is divergent.