Evaluating $\sum_{i=a+1}^{N}\frac{i(i-1)}{i-a}$

Question

I am trying to solve the German Tank Problem. There might be numerous ways of finding the expected value of N. However, the way in which I am proceeding, I need to find this sum. However I am stuck at it.

$$\sum_{i=a+1}^{N}\frac{i(i-1)}{i-a}$$

Any help will be greatly appreciated.

Answer 1

We have

\begin{align}\sum_{i = a+1}^N \frac{i(i-1)}{i-a} &= \sum_{i = a+1}^N \frac{((i - a) + a)(i - 1)}{i - a} \\ &= \sum_{i = a+1}^N \left(1 + \frac{a}{i - a}\right)(i - 1)\\ &= \sum_{i = a+1}^N \left(1 + \frac{a}{i - a}\right)((i - a) + (a - 1))\\ &= \sum_{i = a+1}^N \left[((i - a) + a) + \left(a - 1 + \frac{a(a-1)}{i - a}\right)\right]\\ &= \sum_{i = a+1}^N (i + a - 1) + \sum_{i = a+1}^N \frac{a(a-1)}{i-a}\\ &= \sum_{i = 1}^{N-a} (i + 2a - 1) + \sum_{i = 1}^{N-a} \frac{a(a-1)}{i}\\ &= \frac{(N-a)(N-3a-1)}{2} + a(a-1)H_{N-a} \end{align}

where $H_k$ denotes the $k$ harmonic number.

Answer 2

Mathematica 9.0 gives:

$$(1/2)(a^2(-3 + 2\gamma) + N(N-1 + N) + a(1 - 2\gamma + 2N) + 2a(a-1)\psi(0, 1 - a + N))$$

Where $\gamma$ is the Euler gamma constant and $\psi$ is the polygamma function.