1. The series sums to 4/3.

1. The powers of 2 in the denominators suggest trying x = 1/2. Then we see the numerical series is x = 1/2 substituted into 1 + x² + x⁴ + x⁶ + x⁸ + … We recognize this as the series expansion for 1/(1 – x²). Then, 1 + 1/2² + 1/2⁴ + 1/2⁶ + 1/2⁸ + … = 1/(1 – (1/2)²) = 4/3.

1. The series sums to 27/32.

1. The powers of 3 in the denominators suggest trying x = 1/3. Then we see the numerical series is x = 1/3 substituted into 2x + 4x³ + 6x⁵ + 8x⁷ + 10x⁹ + …The relation between the coefficient and the exponent of each term suggests viewing this as a derivative: 2x + 4x³ + 6x⁵ + 8x⁷ + 10x⁹ + … = (x² + x⁴ + x⁶ + x⁸ + x¹⁰ + …)’  Of course, we can add any constant to the series being differentiated, without changing its derivative. Adding 1 gives a familiar series. 2x + 4x³ + 6x⁵ + 8x⁷ + 10x⁹ + … = (1 + x² + x⁴ + x⁶ + x⁸ + x¹⁰ + …)’ = (1/(1 – x²))’ = 2x/(1 – x²)² Substituting in x = 1/3, we find 2/3 + 4/3³ + 6/3⁵ + 8/3⁷ + 10/3⁹ + … = (2/3)/(1 – 1/3²)² = 27/32

1. The series sums to 16/17.

1. The powers of 4 in the denominators suggest trying x = 1/4. Then we see the numerical series is x = 1/4 substituted into 1 – x² + x⁴ – x⁶ + x⁸ + …We recognize this as the series expansion for 1/(1 + x²). Then 1 – 1/4² + 1/4⁴ – 1/4⁶ + 1/4⁸ – … = 1/(1 + 1/4²) = 16/17.

1. The series sums to π/4.

1. Here we have fewer clues; no obvious choice for x. But in order to view this as a power series evaluated at some x = a, then the numerical series should contain some number raised to different powers. With this assumption, the only choice for x is x = 1. But what powers should we use? Trying 1, x, x², x³, … gives 1 – x/3 + x²/5 – x³/7 + x⁴/9 – x⁵/11 + …This doesn’t look like a familiar series. If we match the exponent of x to the denominator, then we might recognize this series as the integral of a familiar series.x – x³/3 + x⁵/5 – x⁷/7 + x⁹/9 – x¹¹/11 + 2 + x⁴ – x⁶ + x⁸ – x¹⁰ + …)dx = ∫1/(1 + x²) dx= arctan(x). Substituting x = 1 we find: 1 – 1/3 + 1/5 – 1/7 + 1/9 – 1/11 + …= arctan(1) = π/4.

1. The series sums to π/√3.

1. The previous problem has the same set of denominators as this problem, which saw the exponent of x increasing by 2 between successive terms. The powers of 3 in the numerators suggest taking x = 3, but since the powers of 3 increase by 1 between successive terms of the series, maybe x = √3 is a better choice. Then the series has the form x² – x⁴/3 + x⁶/5 – x⁸/7 + x¹⁰/9 – … This isn’t exactly a series we recognize, but it’s close. Try factoring out an x (NOT an x², because the series from the previous problem, which looks pretty similar to this, starts with an x). x⋅(x – x³/3 + x⁵/5 – x⁷/7 + x⁹/9 – … ) = x⋅arctan(x) where we’ve used the arctan(x) series from the previous problem. Substituting x = √3 we see 3 – 3²/3 + 3³/5 – 3⁴/7 + 3⁵/9 – … = (√3)⋅arctan(√3) = (√3)⋅(π/3) = π/√3.

1. The series is x – x⁵/2 + x⁹/4! – x¹³/6! + … .

1. First, recognize that x⋅cos(x²) is 1/2 the derivative of sin(x²). Substituting u= x² in the series for sin(u), we find sin(x²)= x² – x⁶/3! + x¹⁰/5! – x¹⁴/7! + … Then 2x⋅cos(x²) = (sin(x²))’= (x² – x⁶/3! + x¹⁰/5! – x¹⁴/7! + …)’ = 2x – 6x⁵/3! + 10x⁹/5! – 14x¹³/7! + … and so x⋅cos(x²)= x – x⁵/2 + x⁹/4! – x¹³/6! + … Alternately, we could substitute x² for x in the series expansion of cos(x), them multiply that series by x.

1. The series is 1 + x + 3x²/2 + 7x³/6 + 25x⁴/24 + 27x⁵/40 + 331x⁶/720 +…

1. Substitute u = x + x² in the series e^u = 1 + u + u²/2! + u³/3! + u⁴/4! + … obtaining e^{x + x2} = 1 + (x + x²) + (x + x²)²/2! + (x + x²)³/3! + (x + x²)⁴/4! + …= 1 + (x + x²) + (x² + 2x³ + x⁴)/2! + (x³ + 3x⁴ + 3x⁵ + x⁶)/3! + (x⁴ + 4x⁵ + 6x⁶ + 4x⁷ + x⁸)/4! + …Now group together like powers of x. e^{x + x2} = 1 + x + (1 + 1/2!)x² + (1 + 1/3!)x³ + (1/2! + 3/3! + 1/4!)x⁴ + (3/3! + 4/4! + 1/5!)x⁵ + … = 1 + x + 3x²/2 + 7x³/6 + 25x⁴/24 + 27x⁵/40 + 331x⁶/720 + …

1. The series is 1 + 3x + 6×2 + 10×3 + 15×4 + 21×5 + …

1.  First, recognize that 1/(1 – x)³ is 1/2 the second derivative of 1/(1 – x). Recall the series for 1/(1 – x) is 1/(1 – x) = 1 + x + x² + x³ + x⁴ + x⁵ + … Then (1/(1 – x))’ = 1 + 2x + 3x² + 4x³ + 5x⁴ + 6x⁵ + … (1/(1 – x))” = 2 + 3⋅2x + 4⋅3x² + 5⋅4x³ + 6⋅5x⁴ + 7⋅6x⁵ + …Then the series for 1/(1 – x)³ is:                1 + (3⋅2/2)x + (4⋅3/2)x² + (5⋅4/2)x³ + (6⋅5/2)x⁴ + (7⋅6/2)x⁵ + …= 1 + 3x + 6x² + 10x³ + 15x⁴ + 21x⁵ + …

1. The series is x – 2x³/3 + 2x⁵/15 – 4x⁷/315 + 2x⁹/2835 – 4x¹¹/155925 + …

1. Multiply the series for cos(x) by the series for sin(x) and group together terms having like powers of x. cos(x)⋅sin(x) = (1 – x²/2! + x⁴/4! – x⁶/6! + x⁸/8! – …)⋅(x – x³/3! + x⁵/5! – x⁷/7! + …) = x – (1/3! + 1/2!)x³ + (1/5! + (1/2!)⋅(1/3!) + 1/4!)x⁵ – (1/7! + (1/2!)⋅(1/5!) + (1/4!)⋅(1/3!) + 1/6!)x⁷ – (1/9! + (1/2!)⋅(1/7!) + (1/4!)⋅(1/5!) + (1/6!)⋅(1/3!) + 1/8!)x⁹ + ..

1. The series is x + x⁶/6 + x¹¹/11 + x¹⁶/16 + x²¹/21 + x²⁶/26 + …

1.  The series for 1/(1 – x⁵) can be found by substituting u = x⁵ into the series for 1/(1 – u). 1/(1 – x⁵) = 1 + x⁵ + (x⁵)² + (x⁵)³ + (x⁵)⁴ + (x⁵)⁵ + …= 1 + x⁵ + x¹⁰ + x¹⁵ + x²⁰ + x²⁵ + …Then the integral ∫1/(1 – x⁵) dx has series obtained by integrating this, so ∫1/(1 – x⁵) dx = x + x⁶/6 + x¹¹/11 + x¹⁶/16 + x²¹/21 + x²⁶/26 + …

1.  The series is ∑_n=0^∞a_nxⁿ,where a_n = (-1)ⁿ(n+1)(n+2)/2

1. For f(x) = 1/(1 + x)³ we compute some derivatives and seek a pattern. f'(x)= -3/(1 + x)⁴ f”(x)= 3⋅4/(1 + x)⁵ f”'(x)= -3⋅4⋅5/(1 + x)⁶ and so on. Evaluating the function and its derivatives at x = 0 we find f(0)= 1 f'(0)= -3 f”(0)= 3⋅4 f”'(0)= -3⋅4⋅5 and so on. Then the coefficients of the Taylor series, f(n)(0)/n!, are f(0) = 1 f'(0)/1! = -3 f”(0)/2! = 3⋅4/2! = 3⋅4/2 f”'(0)/3! = -3⋅4⋅5/3! = – 4⋅5/2 We need a few more terms to find the pattern. f(⁴)(0)/4! = 3⋅4⋅5⋅6/4! = 5⋅6/2 f(⁵)(0)/5! = -3⋅4⋅5⋅6⋅7/5! = -6⋅7/2 Now we see it: f(ⁿ)(0)/n! = (-1)ⁿ(n+1)(n+2)/2

1. The expansion is f(x)= 1 -2(x-1) -3(x-1)² – (x-1)³.

1. We compute the derivatives. Only a few are nonzero, so we need to expend much energy looking for the pattern. f(1)= 1 f'(1)= -2 f”(1)= -6 f”'(x)= -6 f(ⁿ)(x) = 0 for all n > 3. Then the Taylor series is f(x) = 1 – (2/1!)(x-1) – (6/2!)(x-1)² – (6/3!)(x – 1)³ = 1 -2(x-1) -3(x-1)² – (x-1)³

1. The series is x³ + x⁴ + x⁵/2! + x⁶/3! + x⁷/4! + … + xⁿ⁺³/n! + …

1. Recall the Taylor series for e^x= 1 + x + x²/2! + x³/3! + x⁴/4! + … Multiplying by x³ raises all the exponents by 3: x³e^x= x³ + x⁴ + x⁵/2! + x⁶/3! + x⁷/4! + … + xⁿ⁺³/n! + …

1. sin²(x)= -x² + 2³x⁴/4! – 2⁵x⁶/6! + …

1. Recall sin²(x) = (1 – cos(2x))/2. The Taylor series for cos(2x) is cos(2x)= 1 – (2x)²/2! + (2x)⁴/4! – (2x)⁶/6! + ….So 1 – cos(2x) has series expansion 1 – cos(2x)= -2²x²/2! + 2⁴x⁴/4! – 2⁶x⁶/6! + …and we find sin²(x)= -x² + 2³x⁴/4! – 2⁵x⁶/6! + …

1. The limit is -1/6.

1. Substitute in the Taylor series for sin(x), obtaining sin(x) – x = (x -x³/3! + x⁵/5! – …) – x= -x³/3! + x⁵/5! – … and so (sin(x) – x)/x³ = -1/3! + x²/5! – … where all the omitted terms have powers of x higher than 2, specifically, x⁴, x⁶, x⁸ and so on. As x → 0, only the -1/3! remains.

1. The radius of convergence is 1, the interval of convergence is [-1,1).

1. Apply the Ratio Test lim_{n → ∞}|(xn+1/√(n+1))/(xⁿ/√n)| = lim_{n → ∞} |x|⋅√(n/(n+1)) = |x| By the Ratio Test, the series converges for |x| < 1, that is, -1 < x < 1. The radius of convergence is 1. At the left endpoint, the series becomes ∑_n=1^∞(-1)ⁿ/√n. This converges by the Alternating Series Test. At the right endpoint, the series becomes ∑_n=1^∞1/√n. This is a p-series with p = 1/2, so diverges. The interval of convergence is [-1,1).

1. The radius of convergence is 5, the interval of convergence is [-4,6).

1. Apply the Ratio Test lim_{n → ∞}|((x-1)ⁿ⁺¹/((n+1)5ⁿ⁺¹) / ((x-1)ⁿ/(n 5ⁿ)| = lim_{n → ∞} |(x-1)ⁿ⁺¹/(x-1)ⁿ|⋅(n/(n+1))⋅(5ⁿ/5ⁿ⁺¹) = |x-1|/5 By the Ratio Test, the series converges for |x-1|/5 < 1, that is for -5 < x-1 < 5, or -4 < x < 6. The radius of convergence is 5. At the left endpoint the series becomes ∑_n=1^∞(-5)ⁿ/(n 5ⁿ) = ∑_n=1^∞(-1)ⁿ/n convergent by the Alternating Series Test. At the right endpoint the series becomes ∑_n=1^∞(5)ⁿ/(n 5ⁿ)= ∑_n=1^∞ 1/n This is the harmonic series, hence divergent. The interval of convergence is [-4, 6).

1. The radius of convergence is 2, the interval of convergence is (-2,2].

1. Apply the Ratio Test lim_{n → ∞}|((-1)ⁿ⁺¹xⁿ⁺¹)/(2ⁿ⁺¹ln(n+1)) / ((-1)ⁿxⁿ)/2nln(n))| = lim_{n → ∞} |-(xn+1/xn)|⋅(2n/2n+1)⋅(ln(n)/ln(n+1)) = |-x|/2 where limn → ∞ln(n)/ln(n+1) = 1 is obtained by applying l’Hôpital’s rule to lim_{x → ∞}ln(x)/ln(x+1) By the Ratio Test, the series converges for |-x|/2 < 1, that is, for -2 < x < 2. The radius of convergence is 2. At the left endpoint, the series becomes ∑_n=2^∞(-1)ⁿ(-2)ⁿ/(2ⁿln(n))= ∑_n=2^∞ 2ⁿ/(2ⁿ ln(n))) = ∑_n=2^∞1/ln(n) Now ln(n) < n, so 1/ln(n) > 1/n and the series diverges by comparison with the harmonic series. At the right endpoint, the series becomes ∑n=2∞(-1)n2n/(2nln(n)) = ∑_n=2^∞ (-1)ⁿ/ln(n) converging by the Alternating Series Test. The interval of convergence is (-2,2].

1. The radius of convergence is 1/3, the interval of convergence is [-2/3, 0].

1. Apply the Ratio Test lim_{n → ∞}|((3x+1)ⁿ⁺¹/(n+1)²) / ((3x+1)ⁿ/n²)|= lim_{n → ∞} |(3x+1)ⁿ⁺¹/(3x+1)ⁿ|⋅(n²/(n+1)²)= |3x+1| By the Ratio Test, the series converges for |3x+1| < 1, that is, -1 < 3x+1 < 1, so -2 < 3x < 0, giving -2/3 < x < 0. The radius of convergence is 1/3. At the left endpoint, the series becomes ∑_n=1^∞ (-1)ⁿ/n² convergent by the Alternating Series Test. At the right endpoint, the series becomes ∑_n=1^∞ 1ⁿ/n² convergent, being a p-series with p= 2. The interval of convergence is [-2/3, 0].

1. The radius of convergence is ∞, the interval of convergence is (-∞, ∞).

1. Apply the Ratio Test: lim_{n → ∞}|(xⁿ⁺¹/(n+1)ⁿ⁺¹)/(xⁿ/nⁿ)| = lim_{n → ∞} |xⁿ⁺¹/xⁿ|⋅((nⁿ)/((n+1)ⁿ⁺¹)) = lim_{n → ∞}|x|⋅(n/(n+1))ⁿ⋅(1/(n+1)) = 0 for all x Here we have used lim_{n → ∞}(1 + 1/n)ⁿ= e, we see lim_{n → ∞}(n/(n+1))ⁿ= 1/e to be sure we needn’t be concerned with the factor (n/(n+1))ⁿ in the limit. By the Ratio Test, this series converges for al x, so the radius of convergence is ∞ and the interval of convergence is (-∞, ∞).

1. The series is -1 – x² – x⁴ – x⁶ – x⁸-…

1. First, note that 1/(x² – 1)= -1/(1 – x²). Now apply the series of Example 3 to conclude 1/(x² – 1)= -(1 + x² + x⁴ + x⁶ + x⁸ + …)

    1. Adding the series for (1/2)(1/(x+1)) and for (1/2)(1/(x-1)), and rearranging the terms, we           obtain the series from problem 1.

    1. From Example 1 we know a series for 1/(1 + x) is 1 – x + x² – x³ + x⁴– … The series for 1/(x – 1)= -(1/(1 – x)) is -1 – x – x² – x³ – x⁴ – … Next, note that using a partial fractions expansion we have 1/(x² – 1) = (1/2)(1/(x-1)) – (1/2)(1/(x+1)) Assuming we can add the series and rearrange them in any way we wish (This will be addressed in the next section), we find 1/(x² – 1)= (1/2)(-1 – x – x² – x³ – x⁴ – …) – (1/2)(1 – x + x² – x³ + x⁴ – …) = -1 – x² – x⁴; – x² – … agreeing with       the solution to problem 1.

    1. The series is (1/2 – 1/3) – (1/4 – 1/9)x + (1/8 – 1/27)x² – (1/16 – 1/81)x³ + …This series is valid                for |x| < 2.

    1. Using the method of partial fractions, we find: 1/(x² + 5x + 6) = 1/((x + 2)(x + 3)) = 1/(x + 2) – 1/(x + 3). Using the method of Example 4 we find these power series: 1/(2 + x) = 1/(2(1 + x/2)) = (1/2)(1 – x/2 + x²/4 – x³/8 + x⁴/16 – …) 1/(3 + x) = 1/(3(1 + x/3)) = (1/3) (1 – x/3 + x²/9 – x³/27 + x⁴/81 – …) Subtracting these we find the series for 1/(x² + 5x + 6) is: (1/2 – 1/3) – (1/4 –         1/9)x + (1/8 – 1/27)x² – (1/16 – 1/81)x³ + … The series for 1/(2 + x) is valid for: |x/2| < 1, that is, |x| < 2. The series for 1/(3 + x) is valid for |x/3| < 1, that is, |x| < 3. Both expansions are valid for the intersection of these intervals, |x| < 2.

    1. The series is (1/2 + 1/3) – (1/4 + 1/9)x + (1/8 + 1/27)x² – (1/16 + 1/81)x³ + …This expansion is            valid for |x| < 2.

    1. Using the method of partial fractions, we find (2x + 5)/(x² + 5x + 6) = (2x + 5)/((x + 2)(x + 3))= 1/(x + 2) + 1/(x + 3) Recalling the series for 1/(2 + x) and 1/(3 + x) from problem 3, we find the series for (2x + 5)/(x² + 5x + 6) is the sum of these (1/2 + 1/3) – (1/4 + 1/9)x + (1/8 + 1/27)x² – (1/16 + 1/81)x³ + … As in problem 3, this expansion is valid for |x| < 2.

    1. The series is 1 + (x + 1) + (x + 1)² + (x + 1)³ + (x + 1)⁴ + … The series is valid for -2 < x < 0.

    1. The expansion for 1/(1 – (x + 1)) is 1 + (x + 1) + (x + 1)² + (x + 1)³ + (x + 1)⁴ + … This converges          for |x + 1| < 1. That is, -1 < x + 1 < 1, or -2 < x < 0