Summation rule index In the simplification shown below, The index are changed but when i follow the index change rule, i cant find the same final answer.
$$\sum^{n-1}_{i=0} \frac{n}{n-i}=n\sum^{n}_{i=1} \frac{1}{i}$$
Thanks

Write it as
$$\sum^{n-1}_{i=0} \frac{n}{n-i}= \frac{n}{n} + \frac{n}{n-1} + \dots + \frac{n}{1}$$
Read from the end to the beginning.
$$\frac{n}{1} + \frac{n}{2} + \dots + \frac{n}{n} = n \Big(\frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{n}\Big) = n\sum^{n}_{i=1} \frac{1}{i}$$
