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$$\sum_{n=1}^k \frac{1}{1+na}$$

Where $0\lt a \lt 1$ and $k\gt 0\in\Bbb{Z}$

I'm curious because I want to calculate the present value of an annuity with simple interest, but I don't know the summation of this series, or how to figure it out. Any help would be great.

Collin Angell
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    Since it is close to the harmonic sum it is easy to find a closed form for the asymptotics, but I don't think there's a closed form for partial sums – Evariste Aug 12 '21 at 17:21
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    There is no simple closed form, but you can write the sum in terms of the digamma function: $$ \sum\limits_{n = 1}^k {\frac{1}{{1 + na}}} = \frac{1}{a}\sum\limits_{n = 1}^k {\frac{1}{{1/a + n}}} = \frac{{\psi (k + 1/a + 1) - \psi (1/a + 1)}}{a}. $$ See https://en.wikipedia.org/wiki/Digamma_function – Gary Aug 12 '21 at 17:29
  • Thanks! I'll have to read up on the digamma function. – Collin Angell Aug 13 '21 at 19:32

2 Answers2

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Note that when $a=1$, these are the partial sums of the harmonic series, minus the first term: $$\sum_{n=1}^{k} \frac{1}{1+n} = \sum_{m=2}^{k+1} \frac{1}{m}.$$

It is my understanding that there is not a "nice" closed form (e.g., in terms of rational functions of $n$) for these partial sums; see this question for more discussion, particularly the accepted answer. Since there is not a "nice" closed form for this specific case, there isn't going to be one for your more general expression.

J.E.
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$$S_k=\sum_{n=1}^k \frac{1}{1+na}=\frac{\psi \left(k+1+\frac{1}{a}\right)-\psi \left(1+\frac{1}{a}\right)}{a}=\frac{H_{k+\frac{1}{a}}-H_{\frac{1}{a}}}{a}$$ If $k$ is large $$S_k=\frac{-H_{\frac{1}{a}}+\log (k)+\gamma }{a}+\frac{a+2}{2 a^2 k}-\frac{a^2+6a+6 }{12 a^3 k^2}+O\left(\frac{1}{k^3}\right)$$ and, if you want to solve for $k$ the equation $S_k=b$, using the expansion to $O\left(\frac{1}{k^2}\right)$, you would have an approximate solution given by $$k \sim -\frac {2+a} {2a\,W(t)}\quad \text{where} \qquad t=-\frac {2+a} {2a} \exp\Big[\gamma-a b-H_{\frac{1}{a}} \Big]$$ where $W(t)$ is Lambert function.

Claude Leibovici
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