Kaitlin Jacobson

Answered

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 $S=\sum _{i=1}^{100}\frac{100\cdot 99\cdot ...\cdot (100-i+1)}{{100}^{i}}\cdot i$

$=\sum _{i=1}^{100}\frac{{}^{100}{P}_{i}}{{100}^{i}}\cdot i$

$\frac{100}{100}\cdot 1+\frac{100\cdot 99}{{100}^{2}}\cdot 2+\frac{100\cdot 99\cdot 98}{{100}^{3}}\cdot 3+\frac{100\cdot 99\cdot 98\cdot 97}{{100}^{4}}\cdot 4+\dots +\frac{100\cdot 99\cdot 98\cdot \dots \cdot 1}{{100}^{100}}$

What I have tried:

Above sum can be written as $S=\sum _{i=1}^{100}\frac{100\cdot 99\cdot ...\cdot (100-i+1)}{{100}^{i}}\cdot i$

$=\sum _{i=1}^{100}\frac{{}^{100}{P}_{i}}{{100}^{i}}\cdot i$

Answer & Explanation

perdrigatil5

Expert

2022-12-20Added 2 answers

The sum can be written as

$\sum _{i=0}^{100}\frac{100!i}{(100-i)!\cdot {100}^{i}}$

$=100!\sum _{i=0}^{100}\frac{100-i}{i!\cdot {100}^{100-i}}$

$=\frac{100!}{{100}^{100}}\sum _{i=0}^{100}\frac{(100-i)\cdot {100}^{i}}{i!}$

$=\frac{100!}{{100}^{100}}\sum _{i=0}^{100}(\frac{{100}^{i+1}}{i!}-\frac{{100}^{i}}{(i-1)!})$

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 $\sum _{i=0}^{100}f(i)$ simply tells us to sum the terms over the natural numbers ranging from 0 to 100. When you consider $\sum _{i=0}^{100}f(100-i)$, you can see that it gives you the same set of terms, (i.e. $f(100),f(99),\cdots ,f(0)$). 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 $f(i+1)-f(i)$. Let us define $f(i)=\frac{{100}^{i}}{(i-1)!}$

So we have to evaluate

$\sum _{i=0}^{100}(f(i+1)-f(i))$

$=f(1)-f(0)+f(2)-f(1)+f(3)-f(2)+\cdots +f(100)-f(99)+f(101)-f(100)$

After cancelling, we have left $f(101)-f(0)$. Now recall

$f(i)=\frac{{100}^{i}}{(i-1)!}=\frac{{100}^{i}\cdot i}{i!}$

Then $f(101)={\displaystyle \frac{{100}^{101}\cdot 101}{101!}}$ and $f(0)=0$

Thus, the sum is

$\frac{100!}{{100}^{100}}\times \frac{{100}^{101}\cdot 101}{101!}$

$=\overline{){\displaystyle 100}}$

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.

$\sum _{i=0}^{100}\frac{100!i}{(100-i)!\cdot {100}^{i}}$

$=100!\sum _{i=0}^{100}\frac{100-i}{i!\cdot {100}^{100-i}}$

$=\frac{100!}{{100}^{100}}\sum _{i=0}^{100}\frac{(100-i)\cdot {100}^{i}}{i!}$

$=\frac{100!}{{100}^{100}}\sum _{i=0}^{100}(\frac{{100}^{i+1}}{i!}-\frac{{100}^{i}}{(i-1)!})$

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 $\sum _{i=0}^{100}f(i)$ simply tells us to sum the terms over the natural numbers ranging from 0 to 100. When you consider $\sum _{i=0}^{100}f(100-i)$, you can see that it gives you the same set of terms, (i.e. $f(100),f(99),\cdots ,f(0)$). 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 $f(i+1)-f(i)$. Let us define $f(i)=\frac{{100}^{i}}{(i-1)!}$

So we have to evaluate

$\sum _{i=0}^{100}(f(i+1)-f(i))$

$=f(1)-f(0)+f(2)-f(1)+f(3)-f(2)+\cdots +f(100)-f(99)+f(101)-f(100)$

After cancelling, we have left $f(101)-f(0)$. Now recall

$f(i)=\frac{{100}^{i}}{(i-1)!}=\frac{{100}^{i}\cdot i}{i!}$

Then $f(101)={\displaystyle \frac{{100}^{101}\cdot 101}{101!}}$ and $f(0)=0$

Thus, the sum is

$\frac{100!}{{100}^{100}}\times \frac{{100}^{101}\cdot 101}{101!}$

$=\overline{){\displaystyle 100}}$

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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