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Tristian Velazquez

Tristian Velazquez

Answered question

2022-06-27

I need to find the values of a R + for which the following sum converges.
k = 1 a n = 1 k 1 n

Answer & Explanation

Sawyer Day

Sawyer Day

Beginner2022-06-28Added 30 answers

We compute
lim k + k ( a n = 1 k 1 n a n = 1 k + 1 1 n 1 ) = lim k + k ( a 1 k + 1 1 ) = lim k + k ( e 1 k + 1 log a 1 ) = log a .
By Raabe's test the series converges when log a > 1, i.e., 0 < a < e 1 and diverges when a > e 1 . To check the convergence for a = e 1 , we use n = 1 k 1 n < log k + 1, to find
k = 1 e n = 1 k 1 n e 1 k = 1 e log k = e 1 k = 1 1 k .
But the harmonic series diverges, hence the series diverges when a = e 1

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