Why is $\sum\limits_{i = 1}^n \frac{n}{n-i+1}$ equal to $n \sum\limits_{i = 1}^n \frac{1}{i}$?

Question

Assume we have the sum $\sum\limits_{i = 1}^n \frac{n}{n-i+1}$ why is this equal to $n \sum\limits_{i = 1}^n \frac{1}{i}$?

Answer 1

$$\sum_{i = 1}^n \frac{n}{n-i+1}=n\sum_{i = 1}^n \frac{1}{n-i+1}=n \left(\frac{1}{n}+\frac{1}{n-1}+...+\frac{1}{1}\right)$$

Answer 2

The first comment above gives an excellent reason, but you can formalize it as follows: substitute $\;k:=n-i+1\;$ , then this is an ascending function of $\;n\;$ (differentiate, for example...or directly), and we get that

$$i=1\implies k=n\;,\;\;i=n\implies k=1$$

so we in fact get

$$\sum_{i=1}^n(n-i+1)=\sum_{k=1}^nk$$

and in the original question:

$$\sum_{i=1}^n\frac n{n-i+1}=n\sum_{i=1}^n\frac1{n-i+1}=n\sum_{k=1}\frac1k$$

Answer 3

If you simply open the summation you see that:

$$\sum_{i=1}^n\frac n{n-i+1}=\frac n{n}+\frac n{n-1}+\frac n{n-2}+\cdots+\frac n{1}$$$$=n\bigg(\frac 1{n}+\frac 1{n-1}+\frac 1{n-2}+\cdots+\frac 1{1}\bigg)$$ $$=n\bigg(\frac 1{1}+\cdots+\frac 1{n-2}+\frac 1{n-1}+\frac 1{n}\bigg)$$ $$=n\sum_{i=1}^n\frac 1{i}$$