Find the value a for which e^a + e^(2a) + e^(3a) + e^(4a) + · · · = 4. Here e is the Euler’s constant

Question

Find the value a for which e^a + e^(2a) + e^(3a) + e^(4a) + · · · = 4. Here e is the Euler’s constant

Answer 1

The series is geometric with first term e^a and common ratio e^a, so (assuming e^a < 1) the series converges to e^a/(1 - e^a). Setting e^a/(1 - e^a) = 4 and cross multiplying, e^a = 4(1 - e^a). Thus 5e^a = 4, or e^a = 4/5. Hence a = ln(4/5).