Coupon Collector's Problem image of expectation
I'm reading a book to help with a college course. This is the example used in the lecture however I'm not sure how the limit can change in this example. I have followed other youtube videos on the change but the use of i and n here are confusing me. I understand taking the n out and multiplying by the whole summation, its bottom line of (n-i) becoming (i) that has me lost.
We first use $\sum_{i=0}^{n-1}\frac{n}{n-i}=n\sum_{i=0}^{n-1}\frac{1}{n-i}$. So we just need to show that $\sum_{i=0}^{n-1}\frac{1}{n-i}=\sum_{i=1}^n \frac{1}{i}$. I find it simpler to use a different letter (as we do with $u$ substitution in calculus). Let $j=n-i$, so $\frac{1}{n-i}=\frac{1}{j}$. This gives us the transformation of the summand. Now we need to transform the endpoints. When $i=0$, $j=n-0=n$ which is the upper limit on $j$. When $i=n-1$, $j=n-(n-1)=1$, which is the lower limit on $j$. So $$\sum_{i=0}^{n-1}\frac{1}{n-i}=\sum_{j=1}^n \frac{1}{j}.$$ Now if we want, we can replace each $j$ with an $i$ again.