Trouble with a double summation

[docnet]

Homework Statement

verify the last equality

Relevant Equations

verify the last equality

the inner sum is just adding 1/365 n-i number of times. so $\frac{(n-i)}{365}$

the outer sum adds over the index i, so I thought the expression is equal to $\frac{(n-1)n-(n-1)!}{365}$ but it's obviously not equal to that. where did I go wrong?

 

[PeroK]

Homework Statement:: verify the last equality
Relevant Equations:: verify the last equality

View attachment 291798

the inner sum is just adding 1/365 n-i number of times. so $\frac{(n-i)}{365}$

the outer sum adds over the index i, so I thought the expression is equal to $\frac{(n-1)n-(n-1)!}{365}$ but it's obviously not equal to that. where did I go wrong?

If $i = 1$ there are $n-1$ terms, if $i =2$ there are $n -2$ terms etc. Hence there are $1 +2 \dots + n -1$ terms in total.

Do an example with $n = 4$ say, and write out the sum.

 

[docnet]

Homework Statement:: verify the last equality
Relevant Equations:: verify the last equality

View attachment 291798

the inner sum is just adding 1/365 n-i number of times. so $\frac{(n-i)}{365}$

the outer sum adds over the index i, so I thought the expression is equal to $\frac{(n-1)n-(n-1)!}{365}$ but it's obviously not equal to that. where did I go wrong?

If $i = 1$ there are $n-1$ terms, if $i =2$ there are $n -2$ terms etc. Hence there are $1 +2 \dots + n -1$ terms in total.

Do an example with $n = 4$ say, and write out the sum.

If $n=4$, then the sum is

$1/365\cdot \sum^3_{i=1}\sum^4_{j=i+1}1=1/365\cdot (3+2+1)=\frac{6}{365}$.

check:
$\frac{n(n-1)}{2*365}=\frac{4(3)}{2*365}=\frac{6}{365}$

so I realized that I forgot the meaning of the factorial.. oope. I somehow thought it was n! = n+(n-1)+...+2+1 . oh lord.

following the same thought process suggested by @PeroK,

$1/365\cdot sum^{n-1}_{i=1}\sum^n_{j=i+1}1=\frac{(n-1)+(n-2)+...+2+1}{365}$

which is equivalent to $\frac{n(n-1)}{2*365}$ (although I don't know how to prove it, I can see that it is true geometrically.)

 

[anuttarasammyak]

\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}1=\sum_{i=1}^{n-1}(n-i)=n\sum_{i=1}^{n-1}1-\sum_{i=1}^{n-1}i=n(n-1)-\sum_{i=1}^{n-1}i
Then
S:=\sum_{i=1}^{n-1}i=\sum_{i=1}^{n-1}(n-i)
which means just changing order of addition
1+2+...+(n-1)=(n-1)+(n-2)+...+2+1
adding RHS and LHS
2S=\sum_{i=1}^{n-1}n=n\sum_{i=1}^{n-1}1=n(n-1)
This is equivalent to geometrical explanation.

 

[docnet]

S:=\sum_{i=1}^{n-1}i=\sum_{i=1}^{n-1}(n-i)

This confused me at first, but it makes sense after thinking about it. Thank you so much.