Change of indices question.

Question

Suppose we have the series $\sum_{n=1}^{\infty}\frac{cos(nx)}{n}$.

Now suppose we let $k=-n$, hence we get $\sum_{k=-\infty}^{-1}\frac{cos(-kx)}{-k}$ for $k=-1,-2,... .$

Is this correct? I'm a bit confused on how this index change works.

Answer 1

My original post is below but please view the other answer as there are some issues!

Your index is right if a bit awkward. Consider the process one step at a time. If $k=-n$ then by substitution with $n=-k$ the sum becomes: $$\sum_{-k=1}^{\infty}\frac{cos((-k)x)}{(-k)} = \sum_{-k=1}^{\infty}\frac{cos(-kx)}{-k} \textrm{ with} -k=1,2,3, \ldots$$ The bounds now indicate to sum from $1$ to $\infty$ of $-k$. We want to think about $k$ being positive though so we can in a sense multiply each index $k=1,2,3, \ldots$ by $-1$. Now the sum is: $$\sum_{k=-1}^{-\infty}\frac{cos(-kx)}{-k}\textrm{ with } k=-1,-2,-3, \ldots$$ I would stop at this point, since it is clear where to begin such a summation. If you want to write the sum as you did from lowest term to greatest, you indeed get: $$\sum_{k=-\infty}^{-1}\frac{cos(-kx)}{-k}\textrm{ with } k=-1,-2,-3, \ldots$$

Answer 2

Oh! no

In general, in a summation, the transformation of summation index $k$ to $mk$ {scaling) is not allowed even if $m=-1$. One can always shift $k$ as $k+m$. Also, always $$\sum_{k=1}^n f(k)= \sum_{k=n}^{1}f(k).$$ Some time $k=-p$ may work only incidentally as $$ \sum_{k=1}^{n} \frac{\sin k}{k}= \sum_{-n}^{-1} \frac{\sin p}{p}=\sum_{-n}^{-1} \frac{\sin (-p)}{(-p)}$$ You can check that $$\sum_{k=1}^{n} 4k^2 \ne \sum_{p=2}^{2n} p^2.$$ It does not matter if we sum in forward way or backwards. It is always unlike the definite integrals where $$\int_{a}^{b} f(x) dx=-\int_{b}^{a} f(x) dx$$