I'd like to solve this difference equation. Unfortunately, the forcing term is not geometric, so I don't know how to find the solution:
$$ y_{k+1} = y_k + \frac{c}{2k}. $$
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Are you familiar with generating functions? The g.f. $G(x) = \sum_{k \ge 2} y_k x^k$ works well here. – Snowball Oct 18 '13 at 03:51
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@Snowball This expertise is good to have anyway but not necessary in the present case. – Did Oct 18 '13 at 05:47
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Move $y_k$ over to the left side of the equation and take the partial sums on both sides since we conveniently have a telescoping series.
$$\sum \limits_{k=1}^{n-1} (y_{k+1} - y_k) = \sum \limits_{k=1}^{n-1} \frac{c}{2k} ,$$
$$ y_{n} - y_1 = \frac{c}{2} \sum \limits_{k=1}^{n-1} \frac{1}{k} ,$$
where the sum on the right is just the $(n-1)$th harmonic number, $H_{n-1}$. This gives us the solution form,
$$ y_{n} = y_1 + \frac{c}{2} H_{n-1} .$$
David H
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