Question

Evaluate a general solution and prove by mathematical induction:

$\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + ... + \frac{n}{(n+1)!}$

Answer 1

Answer: General solution is $\frac{(n+1)!-1}{(n+1)!}$

Proof:
Let the equation be equal to some sum, $S_n$

Now, we need to find a link between each unit.

$S_n = \frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + ... + \frac{n}{(n+1)!}$

Let's consider a pattern by taking the first four terms.

$S_1 = \frac{1}{2!} = \frac{1}{2}$
$S_2 = \frac{2}{3!} = \frac{1}{3}$
$S_3 = \frac{3}{4!} = \frac{1}{8}$
$S_4 = \frac{4}{5!} = \frac{1}{30}$

Now, let's consider their summations and we'll call them $T_n$.

$T_1 = S_1 = \frac{1}{2}$
$T_2 = T_1 + S_2 = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$
$T_3 = T_2 + S_3 = \frac{5}{6} + \frac{1}{8} = \frac{23}{24}$
$T_4 = T_3 + S_4 = \frac{23}{24} + \frac{1}{30} = \frac{119}{120}$

One aspect of the sequence should be jumping out now; the numerator is always one less than the denominator.

Now, let's consider a pattern with the denominator.
We know that 2! = 2, 3! = 6, 4! = 24, and 5! = 120
We also know that they are the denominators of the sequences.

So, we can generalise our equation as:

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

MI should be fairly simple now, if you have any problems, feel free to message me!