Converting summation ( how do I get from $k \cdot \sum_{j=k}^{2k} \frac{1}{j-k+1}$ to $ k \cdot \sum_{j=1}^{k+1} \frac{1}{j}$)?

Question

I am trying to figure out the steps between these two equal expressions.

$$ k \cdot \sum_{j=k}^{2k} \frac{1}{j-k+1} = k \cdot \sum_{j=1}^{k+1} \frac{1}{j} $$

I get that $2k -k +1 = k+1$, but why does $j=k$ change to $j=1$?

Answer 1

Note that for

• $j$ varing from $k$ up to $2k$

we have that

• $j-k+1$ varies linearly from $1$ to $k+1$

therefore the two sums are equivalent.

If you find confusing using the same $j$ for both sums, let use another index for the second sum, that is for example

$$r=j-k+1 \implies k \cdot \sum_{j=k}^{2k} \frac{1}{j-k+1} = k \cdot \sum_{r=1}^{k+1} \frac{1}{r}$$