# 10.2E: Exercises for Section 10.2

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1) If $$\displaystyle f(x)=\sum_{n=0}^∞\frac{x^n}{n!}$$ and $$\displaystyle g(x)=\sum_{n=0}^∞(−1)^n\frac{x^n}{n!}$$, find the power series of $$\frac{1}{2}\big(f(x)+g(x)\big)$$ and of $$\frac{1}{2}\big(f(x)−g(x)\big)$$.

$$\displaystyle \frac{1}{2}\big(f(x)+g(x)\big)=\sum_{n=0}^∞\frac{x^{2n}}{(2n)!}$$ and $$\displaystyle \frac{1}{2}\big(f(x)−g(x)\big)=\sum_{n=0}^∞\frac{x^{2n+1}}{(2n+1)!}$$.

2) If $$\displaystyle C(x)=\sum_{n=0}^∞\frac{x^{2n}}{(2n)!}$$ and $$\displaystyle S(x)=\sum_{n=0}^∞\frac{x^{2n+1}}{(2n+1)!}$$, find the power series of $$C(x)+S(x)$$ and of $$C(x)−S(x)$$.

In exercises 3 - 6, use partial fractions to find the power series of each function.

3) $$\dfrac{4}{(x−3)(x+1)}$$

$$\displaystyle \frac{4}{(x−3)(x+1)}=\frac{1}{x−3}−\frac{1}{x+1}=−\frac{1}{3(1−\frac{x}{3})}−\frac{1}{1−(−x)}=−\frac{1}{3}\sum_{n=0}^∞\left(\frac{x}{3}\right)^n−\sum_{n=0}^∞(−1)^nx^n=\sum_{n=0}^∞\left((−1)^{n+1}−\frac{1}{3n+1}\right)x^n$$

4) $$\dfrac{3}{(x+2)(x−1)}$$

5) $$\dfrac{5}{(x^2+4)(x^2−1)}$$

$$\displaystyle \frac{5}{(x^2+4)(x^2−1)}=\frac{1}{x^2−1}−\frac{1}{4}\frac{1}{1+\left(\frac{x}{2}\right)^2}=−\sum_{n=0}^∞x^{2n}−\frac{1}{4}\sum_{n=0}^∞(−1)^n\left(\frac{x}{2}\right)^n=\sum_{n=0}^∞\left((−1)+(−1)^{n+1}\frac{1}{2^{n+2}}\right)x^{2n}$$

6) $$\dfrac{30}{(x^2+1)(x^2−9)}$$

In exercises 7 - 10, express each series as a rational function.

7) $$\displaystyle \sum_{n=1}^∞\frac{1}{x^n}$$

$$\displaystyle \frac{1}{x}\sum_{n=0}^∞\frac{1}{x^n}=\frac{1}{x}\cdot \frac{1}{1−\frac{1}{x}}=\frac{1}{x−1}$$

8) $$\displaystyle \sum_{n=1}^∞\frac{1}{x^{2n}}$$

9) $$\displaystyle \sum_{n=1}^∞\frac{1}{(x−3)^{2n−1}}$$

$$\displaystyle \frac{1}{x−3}\cdot \frac{1}{1−\frac{1}{(x−3)^2}}=\frac{x−3}{(x−3)^2−1}$$

10) $$\displaystyle \sum_{n=1}^∞\left(\frac{1}{(x−3)^{2n−1}}−\frac{1}{(x−2)^{2n−1}}\right)$$

Exercises 11 - 16 explore applications of annuities.

11) Calculate the present values $$P$$ of an annuity in which $10,000 is to be paid out annually for a period of 20 years, assuming interest rates of $$r=0.03,\, r=0.05$$, and $$r=0.07$$. Answer $$P=P_1+⋯+P_{20}$$ where $$P_k=10,000\dfrac{1}{(1+r)^k}$$. Then $$\displaystyle P=10,000\sum_{k=1}^{20}\frac{1}{(1+r)^k}=10,000\frac{1−(1+r)^{−20}}{r}$$. When $$r=0.03, \,P≈10,000×14.8775=148,775.$$ When $$r=0.05, \,P≈10,000×12.4622=124,622.$$ When $$r=0.07, \, P≈105,940$$. 12) Calculate the present values $$P$$ of annuities in which$9,000 is to be paid out annually perpetually, assuming interest rates of $$r=0.03,\, r=0.05$$ and $$r=0.07$$.

13) Calculate the annual payouts $$C$$ to be given for 20 years on annuities having present value $100,000 assuming respective interest rates of $$r=0.03,\, r=0.05,$$ and $$r=0.07.$$ Answer In general, $$P=\dfrac{C(1−(1+r)^{−N})}{r}$$ for $$N$$ years of payouts, or $$C=\dfrac{Pr}{1−(1+r)^{−N}}$$. For $$N=20$$ and $$P=100,000$$, one has $$C=6721.57$$ when $$r=0.03; \, C=8024.26$$ when $$r=0.05$$; and $$C≈9439.29$$ when $$r=0.07$$. 14) Calculate the annual payouts $$C$$ to be given perpetually on annuities having present value$100,000 assuming respective interest rates of $$r=0.03, \,r=0.05,$$ and $$r=0.07$$.

15) Suppose that an annuity has a present value $$P=1$$ million dollars. What interest rate $$r$$ would allow for perpetual annual payouts of $50,000? Answer In general, $$P=\dfrac{C}{r}.$$ Thus, $$r=\dfrac{C}{P}=5×\frac{10^4}{10^6}=0.05.$$ 16) Suppose that an annuity has a present value $$P=10$$ million dollars. What interest rate $$r$$ would allow for perpetual annual payouts of$100,000?

In exercises 17 - 20, express the sum of each power series in terms of geometric series, and then express the sum as a rational function.

17) $$x+x^2−x^3+x^4+x^5−x^6+⋯$$ (Hint: Group powers $$x^{3k}, \, x^{3k−1},$$ and $$x^{3k−2}$$.)

$$(x+x^2−x^3)(1+x^3+x^6+⋯)=\dfrac{x+x^2−x^3}{1−x^3}$$

18) $$x+x^2−x^3−x^4+x^5+x^6−x^7−x^8+⋯$$ (Hint: Group powers $$x^{4k}, \, x^{4k−1},$$ etc.)

19) $$x−x^2−x^3+x^4−x^5−x^6+x^7−⋯$$ (Hint: Group powers $$x^{3k}, \, x^{3k−1}$$, and $$x^{3k−2}$$.)

$$(x−x^2−x^3)(1+x^3+x^6+⋯)=\dfrac{x−x^2−x^3}{1−x^3}$$

20) $$\displaystyle \frac{x}{2}+\frac{x^2}{4}−\frac{x^3}{8}+\frac{x^4}{16}+\frac{x^5}{32}−\frac{x^6}{64}+⋯$$ (Hint: Group powers $$\left(\dfrac{x}{2}\right)^{3k}, \, \left(\dfrac{x}{2}\right)^{3k−1},$$ and $$\left(\dfrac{x}{2}\right)^{3k−2}$$.)

In exercises 21 - 24, find the power series of $$f(x)g(x)$$ given $$f$$ and $$g$$ as defined.

21) $$\displaystyle f(x)=2\sum_{n=0}^∞x^n,g(x)=\sum_{n=0}^∞nx^n$$

$$a_n=2, \, b_n=n$$ so $$\displaystyle c_n=\sum_{k=0}^nb_ka_{n−k}=2\sum_{k=0}^nk=(n)(n+1)$$ and $$\displaystyle f(x)g(x)=\sum_{n=1}^∞n(n+1)x^n$$

22) $$\displaystyle f(x)=\sum_{n=1}^∞x^n,\; g(x)=\sum_{n=1}^∞\frac{1}{n}x^n$$. Express the coefficients of $$f(x)g(x)$$ in terms of $$\displaystyle H_n=\sum_{k=1}^n\frac{1}{k}$$.

23) $$\displaystyle f(x)=g(x)=\sum_{n=1}^∞\left(\frac{x}{2}\right)^n$$

$$a_n=b_n=2^{−n}$$ so $$\displaystyle c_n=\sum_{k=1}^nb_ka_{n−k}=2^{−n}\sum_{k=1}^n1=\frac{n}{2^n}$$ and $$\displaystyle f(x)g(x)=\sum_{n=1}^∞n\left(\frac{x}{2}\right)^n$$

24) $$\displaystyle f(x)=g(x)=\sum_{n=1}^∞nx^n$$

In exercises 25 - 26, differentiate the given series expansion of $$f$$ term-by-term to obtain the corresponding series expansion for the derivative of $$f$$.

25) $$\displaystyle f(x)=\frac{1}{1+x}=\sum_{n=0}^∞(−1)^nx^n$$

The derivative of $$f$$ is $$\displaystyle −\frac{1}{(1+x)^2}=−\sum_{n=0}^∞(−1)^n(n+1)x^n$$.

26) $$\displaystyle f(x)=\frac{1}{1−x^2}=\sum_{n=0}^∞x^{2n}$$

In exercises 27 - 28, integrate the given series expansion of $$f$$ term-by-term from zero to $$x$$ to obtain the corresponding series expansion for the indefinite integral of $$f$$.

27) $$\displaystyle f(x)=\frac{2x}{(1+x^2)^2}=\sum_{n=1}^∞(−1)^n(2n)x^{2n−1}$$

The indefinite integral of $$f$$ is $$\displaystyle \frac{1}{1+x^2}=\sum_{n=0}^∞(−1)^nx^{2n}$$.

28) $$\displaystyle f(x)=\frac{2x}{1+x^2}=2\sum_{n=0}^∞(−1)^nx^{2n+1}$$

In exercises 29 - 32, evaluate each infinite series by identifying it as the value of a derivative or integral of geometric series.

29) Evaluate $$\displaystyle \sum_{n=1}^∞\frac{n}{2^n}$$ as $$f′\left(\frac{1}{2}\right)$$ where $$\displaystyle f(x)=\sum_{n=0}^∞x^n$$.

$$\displaystyle f(x)=\sum_{n=0}^∞x^n=\frac{1}{1−x}; \; f′(\frac{1}{2})=\sum_{n=1}^∞\frac{n}{2^{n−1}}=\frac{d}{dx}(1−x)^{−1}\Big|_{x=1/2}=\frac{1}{(1−x)^2}\Big|_{x=1/2}=4$$ so $$\displaystyle \sum_{n=1}^∞\frac{n}{2^n}=2.$$

30) Evaluate $$\displaystyle \sum_{n=1}^∞\frac{n}{3^n}$$ as $$f′\left(\frac{1}{3}\right)$$ where $$\displaystyle f(x)=\sum_{n=0}^∞x^{6n}$$.

31) Evaluate $$\displaystyle \sum_{n=2}^∞\frac{n(n−1)}{2^n}$$ as $$f''\left(\frac{1}{2}\right)$$ where $$\displaystyle f(x)=\sum_{n=0}^∞x^n$$.

$$\displaystyle f(x)=\sum_{n=0}^∞x^n=\frac{1}{1−x}; \; f''\left(\frac{1}{2}\right)=\sum_{n=2}^∞\frac{n(n−1)}{2^{n−2}}=\frac{d^2}{dx^2}(1−x)^{−1}\Big|_{x=1/2}=\frac{2}{(1−x)^3}\Big|_{x=1/2}=16$$ so $$\displaystyle \sum_{n=2}^∞n\frac{(n−1)}{2^n}=4.$$

32) Evaluate $$\displaystyle \sum_{n=0}^∞\frac{(−1)^n}{n+1}$$ as $$\displaystyle ∫^1_0f(t) \, dt$$ where $$\displaystyle f(x)=\sum_{n=0}^∞(−1)^nx^{2n}=\frac{1}{1+x^2}$$.

In exercises 33 - 39, given that $$\displaystyle \frac{1}{1−x}=\sum_{n=0}^∞x^n$$, use term-by-term differentiation or integration to find power series for each function centered at the given point.

33) $$f(x)=\ln x$$ centered at $$x=1$$ (Hint: $$x=1−(1−x)$$)

$$\displaystyle ∫\sum(1−x)^n\,dx=∫\sum(−1)^n(x−1)^n\,dx=\sum \frac{(−1)^n(x−1)^{n+1}}{n+1}$$

34) $$\ln(1−x)$$ at $$x=0$$

35) $$\ln(1−x^2)$$ at $$x=0$$

$$\displaystyle −∫^{x^2}_{t=0}\frac{1}{1−t}dt=−\sum_{n=0}^∞∫^{x^2}_0t^ndx−\sum_{n=0}^∞\frac{x^{2(n+1)}}{n+1}=−\sum_{n=1}^∞\frac{x^{2n}}{n}$$

36) $$f(x)=\dfrac{2x}{(1−x^2)^2}$$ at $$x=0$$

37) $$f(x)=\tan^{−1}(x^2)$$ at $$x=0$$

$$\displaystyle ∫^{x^2}_0\frac{dt}{1+t^2}=\sum_{n=0}^∞(−1)^n∫^{x^2}_0t^{2n}dt=\sum_{n=0}^∞(−1)^n\frac{t^{2n+1}}{2n+1}∣^{x^2}_{t=0}=\sum_{n=0}^∞(−1)^n\frac{x^{4n+2}}{2n+1}$$

38) $$f(x)=\ln(1+x^2)$$ at $$x=0$$

39) $$\displaystyle f(x)=∫^x_0\ln t\,dt$$ where $$\displaystyle \ln(x)=\sum_{n=1}^∞(−1)^{n−1}\frac{(x−1)^n}{n}$$

Term-by-term integration gives $$\displaystyle ∫^x_0\ln t\,dt=\sum_{n=1}^∞(−1)^{n−1}\frac{(x−1)^{n+1}}{n(n+1)}=\sum_{n=1}^∞(−1)^{n−1}\left(\frac{1}{n}−\frac{1}{n+1}\right)(x−1)^{n+1}=(x−1)\ln x+\sum_{n=2}^∞(−1)^n\frac{(x−1)^n}{n}=x\ln x−x.$$

40) [T] Evaluate the power series expansion $$\displaystyle \ln(1+x)=\sum_{n=1}^∞(−1)^{n−1}\frac{x^n}{n}$$ at $$x=1$$ to show that $$\ln(2)$$ is the sum of the alternating harmonic series. Use the alternating series test to determine how many terms of the sum are needed to estimate $$\ln(2)$$ accurate to within $$0.001,$$ and find such an approximation.

41) [T] Subtract the infinite series of $$\ln(1−x)$$ from $$\ln(1+x)$$ to get a power series for $$\ln\left(\dfrac{1+x}{1−x}\right)$$. Evaluate at $$x=\frac{1}{3}$$. What is the smallest $$N$$ such that the $$N^{\text{th}}$$ partial sum of this series approximates $$\ln(2)$$ with an error less than $$0.001$$?

We have $$\displaystyle \ln(1−x)=−\sum_{n=1}^∞\frac{x^n}{n}$$ so $$\displaystyle \ln(1+x)=\sum_{n=1}^∞(−1)^{n−1}\frac{x^n}{n}$$. Thus, $$\displaystyle \ln\left(\frac{1+x}{1−x}\right)=\sum_{n=1}^∞\big(1+(−1)^{n−1}\big)\frac{x^n}{n}=2\sum_{n=1}^∞\frac{x^{2n−1}}{2n−1}$$. When $$x=\frac{1}{3}$$ we obtain $$\displaystyle \ln(2)=2\sum_{n=1}^∞\frac{1}{3^{2n−1}(2n−1)}$$. We have $$\displaystyle 2\sum_{n=1}^3\frac{1}{3^{2n−1}(2n−1)}=0.69300…$$, while $$\displaystyle 2\sum_{n=1}^4\frac{1}{3^{2n−1}(2n−1)}=0.69313…$$ and $$\ln(2)=0.69314…;$$ therefore, $$N=4$$.

In exercises 42 - 45, using a substitution if indicated, express each series in terms of elementary functions and find the radius of convergence of the sum.

42) $$\displaystyle \sum_{k=0}^∞(x^k−x^{2k+1})$$

43) $$\displaystyle \sum_{k=1}^∞\frac{x^{3k}}{6k}$$

$$\displaystyle \sum_{k=1}^∞\frac{x^k}{k}=−\ln(1−x)$$ so $$\displaystyle \sum_{k=1}^∞\frac{x^{3k}}{6k}=−\frac{1}{6}\ln(1−x^3)$$. The radius of convergence is equal to $$1$$ by the ratio test.

44) $$\displaystyle \sum_{k=1}^∞(1+x^2)^{−k}$$ using $$y=\dfrac{1}{1+x^2}$$

45) $$\displaystyle \sum_{k=1}^∞2^{−kx}$$ using $$y=2^{−x}$$

If $$y=2^{−x}$$, then $$\displaystyle \sum_{k=1}^∞y^k=\frac{y}{1−y}=\frac{2^{−x}}{1−2^{−x}}=\frac{1}{2^x−1}$$. If $$a_k=2^{−kx}$$, then $$\dfrac{a_{k+1}}{a_k}=2^{−x}<1$$ when $$x>0$$. So the series converges for all $$x>0$$.

46) Show that, up to powers $$x^3$$ and $$y^3$$, $$\displaystyle E(x)=\sum_{n=0}^∞\frac{x^n}{n!}$$ satisfies $$E(x+y)=E(x)E(y)$$.

47) Differentiate the series $$\displaystyle E(x)=\sum_{n=0}^∞\frac{x^n}{n!}$$ term-by-term to show that $$E(x)$$ is equal to its derivative.

48) Show that if $$\displaystyle f(x)=\sum_{n=0}^∞a_nx^n$$ is a sum of even powers, that is, $$a_n=0$$ if $$n$$ is odd, then $$\displaystyle F=∫^x_0f(t)\, dt$$ is a sum of odd powers, while if $$I$$ is a sum of odd powers, then $$F$$ is a sum of even powers.

49) [T] Suppose that the coefficients an of the series $$\displaystyle \sum_{n=0}^∞a_nx^n$$ are defined by the recurrence relation $$a_n=\dfrac{a_{n−1}}{n}+\dfrac{a_{n−2}}{n(n−1)}$$. For $$a_0=0$$ and $$a_1=1$$, compute and plot the sums $$\displaystyle S_N=\sum_{n=0}^Na_nx^n$$ for $$N=2,3,4,5$$ on $$[−1,1].$$

The solid curve is $$S_5$$. The dashed curve is $$S_2$$, dotted is $$S_3$$, and dash-dotted is $$S_4$$

50) [T] Suppose that the coefficients an of the series $$\displaystyle \sum_{n=0}^∞a_nx^n$$ are defined by the recurrence relation $$a_n=\dfrac{a_{n−1}}{\sqrt{n}}−\dfrac{a_{n−2}}{\sqrt{n(n−1)}}$$. For $$a_0=1$$ and $$a_1=0$$, compute and plot the sums $$\displaystyle S_N=\sum_{n=0}^Na_nx^n$$ for $$N=2,3,4,5$$ on $$[−1,1]$$.

51) [T] Given the power series expansion $$\displaystyle \ln(1+x)=\sum_{n=1}^∞(−1)^{n−1}\frac{x^n}{n}$$, determine how many terms $$N$$ of the sum evaluated at $$x=−1/2$$ are needed to approximate $$\ln(2)$$ accurate to within $$1/1000.$$ Evaluate the corresponding partial sum $$\displaystyle \sum_{n=1}^N(−1)^{n−1}\frac{x^n}{n}$$.

When $$\displaystyle x=−\frac{1}{2}, \;−\ln(2)=\ln\left(\frac{1}{2}\right)=−\sum_{n=1}^∞\frac{1}{n2^n}$$. Since $$\displaystyle \sum^∞_{n=11}\frac{1}{n2^n}<\sum_{n=11}^∞\frac{1}{2^n}=\frac{1}{2^{10}},$$ one has $$\displaystyle \sum_{n=1}^{10}\frac{1}{n2^n}=0.69306…$$ whereas $$\ln(2)=0.69314…;$$ therefore, $$N=10.$$

52) [T] Given the power series expansion $$\displaystyle \tan^{−1}(x)=\sum_{k=0}^∞(−1)^k\frac{x^{2k+1}}{2k+1}$$, use the alternating series test to determine how many terms $$N$$ of the sum evaluated at $$x=1$$ are needed to approximate $$\tan^{−1}(1)=\frac{π}{4}$$ accurate to within $$1/1000.$$ Evaluate the corresponding partial sum $$\displaystyle \sum_{k=0}^N(−1)^k\frac{x^{2k+1}}{2k+1}$$.

53) [T] Recall that $$\tan^{−1}\left(\frac{1}{\sqrt{3}}\right)=\frac{π}{6}.$$ Assuming an exact value of $$\frac{1}{\sqrt{3}})$$, estimate $$\frac{π}{6}$$ by evaluating partial sums $$S_N\left(\frac{1}{\sqrt{3}}\right)$$ of the power series expansion $$\displaystyle \tan^{−1}(x)=\sum_{k=0}^∞(−1)^k\frac{x^{2k+1}}{2k+1}$$ at $$x=\frac{1}{\sqrt{3}}$$. What is the smallest number $$N$$ such that $$6S_N\left(\frac{1}{\sqrt{3}}\right)$$ approximates $$π$$ accurately to within $$0.001$$? How many terms are needed for accuracy to within $$0.00001$$?

$$\displaystyle 6S_N\left(\frac{1}{\sqrt{3}}\right)=2\sqrt{3}\sum_{n=0}^N(−1)^n\frac{1}{3^n(2n+1).}$$ One has $$π−6S_4\left(\frac{1}{\sqrt{3}}\right)=0.00101…$$ and $$π−6S_5\left(\frac{1}{\sqrt{3}}\right)=0.00028…$$ so $$N=5$$ is the smallest partial sum with accuracy to within $$0.001.$$ Also, $$π−6S_7\left(\frac{1}{\sqrt{3}}\right)=0.00002…$$ while $$π−6S_8\left(\frac{1}{\sqrt{3}}\right)=−0.000007…$$ so $$N=8$$ is the smallest $$N$$ to give accuracy to within $$0.00001.$$