Kaitlin Jacobson

2022-12-19

How to evaluate the following sum?
$\frac{100}{100}\cdot 1+\frac{100\cdot 99}{{100}^{2}}\cdot 2+\frac{100\cdot 99\cdot 98}{{100}^{3}}\cdot 3+\frac{100\cdot 99\cdot 98\cdot 97}{{100}^{4}}\cdot 4+\dots +\frac{100\cdot 99\cdot 98\cdot \dots \cdot 1}{{100}^{100}}$
What I have tried:
Above sum can be written as $S=\sum _{i=1}^{100}\frac{100\cdot 99\cdot ...\cdot \left(100-i+1\right)}{{100}^{i}}\cdot i$
$=\sum _{i=1}^{100}\frac{{}^{100}{P}_{i}}{{100}^{i}}\cdot i$

perdrigatil5

The sum can be written as
$\sum _{i=0}^{100}\frac{100!i}{\left(100-i\right)!\cdot {100}^{i}}$
$=100!\sum _{i=0}^{100}\frac{100-i}{i!\cdot {100}^{100-i}}$
$=\frac{100!}{{100}^{100}}\sum _{i=0}^{100}\frac{\left(100-i\right)\cdot {100}^{i}}{i!}$
$=\frac{100!}{{100}^{100}}\sum _{i=0}^{100}\left(\frac{{100}^{i+1}}{i!}-\frac{{100}^{i}}{\left(i-1\right)!}\right)$
which telescopes. Going to the second line from the first, I've substituted 100−i for i.
Edit in response to OP's comments
The substitution does not, in fact, change anything else. The notation $\sum _{i=0}^{100}f\left(i\right)$ simply tells us to sum the terms over the natural numbers ranging from 0 to 100. When you consider $\sum _{i=0}^{100}f\left(100-i\right)$, you can see that it gives you the same set of terms, (i.e. $f\left(100\right),f\left(99\right),\cdots ,f\left(0\right)$). And the sigma notation is used as "counting up", so we do not change the limits.
This is indeed neither an arithmetic series nor a geometric series. Are you familiar with telescoping sums? Notice that the expression inside the summation in the final line has the form $f\left(i+1\right)-f\left(i\right)$. Let us define $f\left(i\right)=\frac{{100}^{i}}{\left(i-1\right)!}$
So we have to evaluate
$\sum _{i=0}^{100}\left(f\left(i+1\right)-f\left(i\right)\right)$
$=f\left(1\right)-f\left(0\right)+f\left(2\right)-f\left(1\right)+f\left(3\right)-f\left(2\right)+\cdots +f\left(100\right)-f\left(99\right)+f\left(101\right)-f\left(100\right)$
After cancelling, we have left $f\left(101\right)-f\left(0\right)$. Now recall
$f\left(i\right)=\frac{{100}^{i}}{\left(i-1\right)!}=\frac{{100}^{i}\cdot i}{i!}$
Then $f\left(101\right)=\frac{{100}^{101}\cdot 101}{101!}$ and $f\left(0\right)=0$
Thus, the sum is
$\frac{100!}{{100}^{100}}×\frac{{100}^{101}\cdot 101}{101!}$
$=\overline{)100}$
Also note that we can change from $\sum _{1}^{100}$ to $\sum _{0}^{100}$ because the term corresponding to i=0 is just 0 and does not change the sum.

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