Change of the summation limits?

Question

I have a sum where the sum limits have changed, and i can't figure out why. It is

N
∑(N!/(k-1)!(N-k)!)*((λ/N)^k)*((1- λ/N)^N-k
K=1
Is the same as
N-1
∑(N!/(u!)(N-u-1)!)*((λ/N)^u+1)*((1- λ/N)^N-u-1
u=0
where u=k-1

I don't understand why the limit on the second sum changes from N to N-1.

Answer 1

Since u = k - 1,

For the lower limit, k = 1. Therefore, u = k - 1 implies
u = 1 - 1 {I'm just substituting k = 1}, which makes u = 0.

For the upper limit, k = n. Therefore, u = k - 1 implies
u = n - 1 {again, substituting k = n}, which makes the upper limit n - 1.

Answer 2

Think of the reverse.

Since u = k - 1, we can replace k by (u + 1) in the FIRST sum.


N
∑(N!/(k-1)!(N-k)!)*((λ/N)^k)*(... λ/N)^N-k
k= u + 1 =1.

Since this is the first summation we should not change the upper limit.

The maximum value for the first summation is k = N.

Or k = u +1 = N. that is u = N-1.

If we have in the summation the figure k we will assign values from 1 to N.

Instead if we have the figure u, then we will assign values from 0 to ( N-1).

Thus when a lower limit is changed accordingly the upper limit is also changing.

Answer 3

Because the limits are for the variable in each case of summation. u is substituted for k-1.

In the first case the sum is from k=1 to N

In the second case the sum is also from k=1 to N, but as the variable in the sum is now u (=k-1), to keep the range the same, the limits on the summation are not from k=1 to N, but from u=0 to N-1 which are equivalent statements as u=k-1.

Answer 4

It might be easier if you think of a different sum. Consider:
N
∑1
k=1
so, for N=1, the sum is 1; for N=23 the sum is 23; etc

then, if you sum from u=0 to N-1 instead of from 1 to N you get the same result.

The same is true for the example you gave, except you have to remember to convert the terms within the sum (which you have done, substituting u-1 for k)

Answer 5

the original lower limit is k=1, since k=u+1 => u+1 = 1 for the lower limit hence u= 1-1= 0
the upper limit is k=N, then u+1 = N => u=N-1

Answer 6

set u = k-1
and substitute :
if k = 1, u = 0
if k = N, u =N-1