Question

The value of $\frac{1}{2!} + \frac{2}{3!} + \dots + \frac{99}{100!}$ is equal to

Answer 1

1 - \frac{1}{100!}

Answer 2

The given series is $S = \frac{1}{2!} + \frac{2}{3!} + \dots + \frac{99}{100!}$.

We can identify the general term of the series. The $n$-th term of the series can be written as $T_n = \frac{n}{(n+1)!}$. The sum is for $n$ ranging from $1$ to $99$.

Let's rewrite the general term $T_n$ by manipulating the numerator: $T_n = \frac{n}{(n+1)!}$ We can write $n$ as $(n+1) - 1$. So, $T_n = \frac{(n+1) - 1}{(n+1)!}$

Now, split the fraction into two parts: $T_n = \frac{n+1}{(n+1)!} - \frac{1}{(n+1)!}$

Simplify the first part: $\frac{n+1}{(n+1)!} = \frac{n+1}{(n+1) \cdot n!} = \frac{1}{n!}$. So, the general term becomes: $T_n = \frac{1}{n!} - \frac{1}{(n+1)!}$

This form is characteristic of a telescoping series. Let's write out the terms of the sum: For $n=1$: $T_1 = \frac{1}{1!} - \frac{1}{2!}$ For $n=2$: $T_2 = \frac{1}{2!} - \frac{1}{3!}$ For $n=3$: $T_3 = \frac{1}{3!} - \frac{1}{4!}$ ... For $n=99$: $T_{99} = \frac{1}{99!} - \frac{1}{100!}$

Now, sum all these terms: $S = T_1 + T_2 + T_3 + \dots + T_{99}$ $S = \left(\frac{1}{1!} - \frac{1}{2!}\right) + \left(\frac{1}{2!} - \frac{1}{3!}\right) + \left(\frac{1}{3!} - \frac{1}{4!}\right) + \dots + \left(\frac{1}{99!} - \frac{1}{100!}\right)$

In this sum, the intermediate terms cancel each other out: The $-\frac{1}{2!}$ from $T_1$ cancels with the $+\frac{1}{2!}$ from $T_2$. The $-\frac{1}{3!}$ from $T_2$ cancels with the $+\frac{1}{3!}$ from $T_3$. This pattern continues until the second-to-last term.

The only terms that remain are the first part of the first term and the second part of the last term: $S = \frac{1}{1!} - \frac{1}{100!}$ Since $1! = 1$, we have: $S = 1 - \frac{1}{100!}$