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 ?
Notation of indexers with multiples in a series
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sequences-and-series
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5You 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. – 2012-03-02
1 Answers
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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,
$\kappa_n=\sigma_{2n}-\frac{\sigma_{n}}{2}=1+\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2n-3}+\frac{1}{2n-1}$