1 Q about Summation

The summation sign means :
k = 0, 1, 2, . . ., n-3, n-2, n-1
You can say that :
k = n-1, n-2, n-3, . . ., 2, 1, 0
Let y = n - k, then the above one can be
y = n - (n-1), n - (n-2), n - (n-3), . . ., n - 2, n - 1, n - 0
that is :
y = 1, 2, 3, . . ., n-2, n-1, n
So the summation can be written as
(from y = 1 to n) [2n - 2(n-y) + 1] / [2n - 2(n-y)]
that is
(from y = 1 to n) [2y + 1] / [2y]
Transfer y to k (dummy variable), so the above one can be written as :
(from k = 1 to n) (2k + 1) / (2k)

For the lower limit, k is changed from 0 to 1 and at the same time the upper limit is changed from (n - 1) to n. That means the number of terms is unchanged but value of k is shifted by 1 unit. So replace k by (k - 1) in the expression will balance out the change, so the new expression is [2n - 2(k - 1) + 1]/[2n - 2(k - 1)] = (2n - 2k + 3)/(2n - 2k + 2) where k from 1 to n.