Change of variables in a sum

Question

If we have the following function $$f(k) = \frac{1}{k-m} \sum_{j=k+1}^{m+n} \frac{1}{j-m}\ \text{where}\ k\in \{m,.....,m+n-1\}$$ How will the function look if we change the variables $i= j-m$ and $r=k-m$?

I have tried to resolve but still having trouble with the final result.

First, since $j\in \{k+1,...m+n \}$ and $i = j-m$ it results that $$ \sum_{j=k+1}^{m+n} \frac{1}{j-m} = \sum_{i=k+1-m}^{n} \frac{1}{i}\ ?$$

Do we have now that $$f(k) = \frac{1}{k-m} \sum_{i=k+1-m}^{n} \frac{1}{i}\ \ ?$$

Also if next we have $r = k-m$ how will the sum change?

$k\in \{m,.....,m+n-1\} => r\in \{0,.....n-1\}$...

Any help is much appreciated, thanks.

Answer 1

Given that $$f(k) = \frac{1}{k-m} \sum_{j=k+1}^{m+n} \frac{1}{j-m}, $$ where $$ k \in \{m,.....,m+n-1\}, $$ and given that $$ i= j-m $$ and $$ r=k-m, $$ we obtain $$ j = i+m $$ and $$ k = r+m, $$ with $$ r \in \{ 0, \ldots, n-1 \}, $$ and hence $$ f(k) = f(r+m) = \frac{1}{r} \sum_{i = k+1-m}^{n} \frac{1}{i} = \frac{1}{r} \sum_{i=r+1}^n \frac{1}{i}, $$ that is, $$ f(r+m) = \frac{1}{r} \sum_{i=r+1}^n \frac{1}{i}, $$ which implies that $$ f(r) = f \big((r-m)+m\big) = \frac{1}{r-m} \sum_{i=r-m+1}^n \frac{1}{i} = \frac{1}{r-m} \sum_{i = r-m+1}^n \frac{1}{i}. $$

Answer 2

Your last sum is $$f(k)=\frac{1}{k-m}\sum_{i=k-m+1}^n\frac 1i$$

with $ r=k-m$, it becomes

$$\displaystyle{f(k)=\frac{1}{r}\sum_{i=r+1}^n\frac 1i}$$

Answer 3

Since the $j$ is the summation index, so this should be transformed You rightly get: $$S=\frac{1}{k-m} \sum_{i=k+1-m}^{n} \frac{1}{i}=\frac{1}{k-m}\left( \sum_{j=1}^{n} \frac{1}{i}- \sum_{i=1}^{k+1-m} \frac{1}{i} \right)=\frac{H_{n}-H_{k+1-m}}{k-m}.$$ where $H_n=1+1/2+1/3+1/4+...+1/n% are called Harmonic numbers, you may see

https://en.wikipedia.org/wiki/Harmonic_number