The number of terms of a
harmonic sequence is even.
The sum of the terms in the odd places (First term + Third Term + Fifth Term + ...and so on) is 2625, and:
The sum of the terms in the even places (Second Term + Fourth Term + Sixth Term + ... and so on) is 4224; and:
Given that the last term exceeds the first by 2205, then identify the terms of the said sequence.
I have been trying to find some analytic solution for this and have not had good results.
I started with an approximation for the partial sum of the classic harmonic series:
H(n) = 1 + 1/2 + 1/3 + ... + 1/n ~= ln(n) + 1/(2n) + 0.5772
[0.5772 is the beginning of the Euler-Mascheroni constant]
Also the partial sums of the alternating harmonic series can be expressed in terms of the classic series (assuming an even number of terms):
A(n) = 1 - 1/2 + 1/3 - 1/4 + ... + 1/(n-1) - 1/n = H(n) - H(n/2)
Initial attempts which made no assumptions about the start or end of the puzzle sequence very quickly got too messy for me to make any progress, so I added the condition that the sequence started at a scaled multiple of 1 in the 1,1/2,1/3,... progression.
Let n be the number of terms in the puzzle sequence and m be the multiplier. Adding the two partial sums then implied:
6849/m ~= ln(n) + 1/(2n) + 0.5772
1599/m ~= ln(2) + 1/(4n)
This reduced to 2.3917 + 0.5708/n ~= ln(n). The closest integer is n=11, which when substituted back yielded m=2234 and m=2267. This is not close to the answer of n=8, m=2520 that Harry found and outright contradicts the third given statement. Back to brainstorming....
A failed attempt
