Sandeep's Harmonic Sums

Algebra Level 3

For each positive integer n n , let H n = 1 1 + 1 2 + ⋯ + 1 n . H_n = \frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{n}. If ∑ n = 4 ∞ 1 n H n H n − 1 = a b \sum_{n=4}^{\infty} \frac{1}{nH_nH_{n-1}} = \frac{a}{b} for relatively prime positive integers a a and b b , find a + b a+b .

This problem is shared by Sandeep S.


The answer is 17.

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Calvin Lin by Calvin Lin

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1 solution

Sandeep Silwal
May 20, 2014

Note that ∑ n = 4 ∞ 1 n H n H n − 1 = ∑ n = 4 ∞ 1 / n H n H n − 1 = ∑ n = 4 ∞ H n − H n − 1 H n H n − 1 = ∑ n = 4 ∞ ( 1 H n − 1 − 1 H n ) = 1 H 3 = 6 11 . \begin{aligned} \sum_{n=4}^{\infty}\frac{1}{nH_nH_{n-1}} &=\sum_{n=4}^{\infty}\frac{1/n}{H_nH_{n-1}} \\ &= \sum_{n=4}^{\infty}\frac{H_n - H_{n-1} }{H_nH_{n-1}} \\ &= \sum_{n=4}^{\infty} \left( \frac{1}{H_{n-1}}-\frac{1}{H_n} \right)\\ &= \frac{1}{H_3} \\ &= \frac{6}{11}. \\ \end{aligned}

Therefore the answer is 6 + 11 = 17 6+11=17 .

30 Helpful 2 Interesting 5 Brilliant 0 Confused

This solution is presented for it's simplicity, as compared to other solutions that were received.

Calvin Lin Staff - 7 years ago

No offence, but I do not think this problem should be a Level 5.

Keshav Tiwari - 6 years, 2 months ago

very elegant solution. Congrat.

Dorina Popescu - 5 years, 4 months ago

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