A Double Euler-Mascheroni Sequence

What is the value of the sum \displaystyle \sum_{i=1}^n \sum_{j=1}^n \frac{1}{i+j}?  Furdui poses this as Question 1.28 in his book Limits, Series, and Fractional Part Integrals [1, p. 5], where he calls the sum a “double Euler-Mascheroni sequence.”  His solution converts the double sum into an expression containing an integral and an infinite series, and then he evaluates those.  In this post I’ll give a different derivation of Furdui’s expression for the value of the double sum – one that uses only standard summation manipulations.

Let \displaystyle H_n denote the nth harmonic number: \displaystyle H_n = \sum_{k=1}^n \frac{1}{k}.  The following lemma is proved in Concrete Mathematics [2, pp. 40-41] and is a straightforward application of summation by parts, but I’ll give a proof here anyway.

Lemma: \displaystyle \sum_{k=1}^n H_k = (n+1) H_{n+1} - (n+1).

Proof:  Swapping the order of summation, we have \displaystyle \sum_{k=1}^n H_k = \sum_{k=1}^n \sum_{j=1}^k \frac{1}{j} = \sum_{j=1}^n \sum_{k=j}^n \frac{1}{j} = \sum_{j=1}^n \frac{n+1-j}{j} = (n+1) \sum_{j=1}^n \frac{1}{j} - \sum_{j=1}^n 1 \\ = (n+1) H_n - n = (n+1) H_n - n + \frac{n+1}{n+1} - 1 = (n+1)H_{n+1} - (n+1).

With the lemma in place, we’re now ready to derive an expression for this double Euler-Mascheroni sequence.

Theorem: \displaystyle \sum_{i=1}^n \sum_{j=1}^n \frac{1}{i+j} = (2n+1) H_{2n+1} - (2n+2) H_{n+1} + 1.

Proof: We have \displaystyle \sum_{i=1}^n \sum_{j=1}^n \frac{1}{i+j} = \sum_{i=1}^n (H_{n+i} - H_i) = \sum_{i=1}^n H_{n+i} - \sum_{i=1}^n H_i = \sum_{i=1}^{2n} H_i - 2 \sum_{i=1}^n H_i.

Applying the lemma, we obtain \displaystyle \sum_{i=1}^n \sum_{j=1}^n \frac{1}{i+j} = (2n+1) H_{2n+1} - (2n+1) - 2 \left( (n+1)H_{n+1} - (n+1)\right) \\ = (2n+1) H_{2n+1} - (2n+2)H_{n+1} + 1.

References

  1. Ovidiu Furdui, Limits, Series, and Fractional Part Integrals, Springer, 2013.
  2. Ronald L. Graham, Donald E. Knuth, and Oren Patashnik, Concrete Mathematics (2nd ed.), Addison-Wesley, 1994.
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