Difference equation: $y_{k+1} = y_{k} + \frac{c}{2k}$

Question

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:

$$ y_{k+1} = y_k + \frac{c}{2k}. $$

Answer 1

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,

$$ y_{n} = y_1 + \frac{c}{2} H_{n-1} .$$