Sigma Notation, to another Form of Sigma Notation and then a final result?!

Question

There is this problem, I fail to understand the solution, $$ \sum_{i=1}^n \frac{i-1}{i!}=? $$

and the solution given is:

Step - 1 : $\sum_{i=1}^n \frac{i}{i!} - \frac{1}{i!}$

Step - 2 : $\sum_{i=1}^n \frac{1}{(i-1)!} - \frac{1}{i!}$

Step - 3 : $1-\frac{1}{n!}$

Now, what i dont understand is, the transition from STEP 1to2 & STEP 2to3

Could you please me give me some explanation.

Thank You, With Respect Umer Selmani

Answer 1

Step 1-2: note that $i! = i(i-1)(i-2)\dots2\cdot 1$

So, $i!\over i $ $=(i-1)(i-2)\dots 2\cdot 1 = (i-1)!$

So, $i\over i!$ $=$$1\over (i-1)!$. This combined with the fact that $\frac ut-\frac vt = {u-v\over t}$ gives us the required result.

For step 2-3, this is a telescopic series. To get a sense of what is being done, write down the terms:

$$\frac 1{0!} - \frac 1{1!} + \frac 1{1!} - \frac 1{2!} \dots -\frac 1{(n-1)!} +\frac 1{(n-1)!} -\frac 1{n!}$$

Note that other than the first and last, every single term that is present’s negative is also present. This leaves $1-\frac 1{n!}$.

For any series $a_n$, $\displaystyle\sum_{s=t}^u a_s - a_{s-1}$ obviously contains both $a_s$ and $-a_{(s+1)-1} = -a_s$ except for where either the $a_{s+1} - a_{s}$ term isn’t in the sum (the term $s=u$) and where the term before $a_s - a_{s-1}$ isn’t in the sum (the term $s = t$).

Answer 2

We have that $\frac{i}{i!}=\frac{i}{i(i-1)!}=\frac{1}{(i-1)!}$ and by telescoping

$$\sum_{i=1}^n \left(\frac{1}{(i-1)!} - \frac{1}{i!}\right)=1\color{red}{-1+1-\frac12+\frac1{2}-\frac16+\ldots+\frac1{(n-1)!}}-\frac1{n!}=1-\frac{1}{n!}$$