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.