Let H_1, H_2, H_3, ... be harmonic numbers defined by H_n = sum_{j=1}^{n} 1/j.

Then, for n = 1, sum_{i=1}^{n} H_i can ...

Question

Let H_1, H_2, H_3, ... be harmonic numbers defined by H_n = sum_{j=1}^{n} 1/j.

Then, for n >= 1, sum_{i=1}^{n} H_i can be expressed as:

A. (n+1)*H_n - n
B. n*H_n
C. (n+1)*H_{n+1} - n
D. n*H_{n+1} - n

Accepted Answer

sum_{i=1}^{n} H_i = (n+1)*H_n - n. The answer is option A. The harmonic number H_n = 1 + 1/2 + 1/3 + ... + 1/n grows like ln(n). When we sum H_1 + H_2 + ... + H_n, we are summing partial sums of 1/j. A standard technique is to reverse the order of summation: each term 1/j appears in H_j, H_{j+1}, ..., H_n, that is, in (n - j + 1) harmonic numbers. This double-counting argument yields sum_{i=1}^{n} H_i = sum_{j=1}^{n} (n - j + 1)/j = (n+1)*H_n - n.

Let H_1, H_2, H_3, ... be harmonic numbers defined by H_n = sum_{j=1}^{n} 1/j. Then, for n >= 1, sum_{i=1}^{n} H_i can be expressed as: A. (n+1)*H_n - n B. n*H_n C. (n+1)*H_{n+1} - n D. n*H_{n+1} - n

Let H_1, H_2, H_3, ... be harmonic numbers defined by H_n = sum_{j=1}^{n} 1/j. Then, for n >= 1, sum_{i=1}^{n} H_i can be expressed as: A. (n+1)H_n - n B. nH_n C. (n+1)H_{n+1} - n D. nH_{n+1} - n