# Notation of indexers with multiples in a series

If $\sigma _{n}=1+\dfrac {1} {2}+\dfrac {1} {3}+\ldots +\dfrac {1} {n}$ what series is given by $\sigma _{2n}$ ? Does that mean we only take the even terms now or does it mean every term is multiplied by 2 ?

-
You add up all the way to the term $\frac{1}{2n}$. But same series, no skipping, no multiplying. It just affects how far you go. –  André Nicolas Mar 2 '12 at 23:44

Since

$$\sigma_{n} = 1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n-1}+\frac{1}{n}$$

is the sum of the reciprocals of $1$ up to $n$, we have that $\sigma_{2n}$ is

$$\sigma_{2n} = 1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n-1}+\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n-1}+\frac{1}{2n}$$

That is, we sum up to $2n$.

If we want to sum only even numbers, we'd have to change our notation and maybe write

$$\omega_n=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2n-2}+\frac{1}{2n}=\frac{\sigma_{n}}{2}$$ and for odd numbers, put,

$$\kappa_n=\sigma_{2n}-\frac{\sigma_{n}}{2}=1+\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2n-3}+\frac{1}{2n-1}$$

-