# Mathematical Induction (2)

• May 11th 2008, 05:48 AM
Tangera
Mathematical Induction (2)
Q: Given that f(n) = (1/2!) + (2/3!) + 3/4!) +...+ [n/(n+1)!], where n is a positve integer, find the values of f(1), f(2) and f(3). hence make a conjecture for a forumla for f(n) in terms of n and prove it using mathematical induction.

• May 11th 2008, 06:21 AM
flyingsquirrel
Hi

$f(1)=\boxed{\frac{1}{2}}$ (note that $2!=2$)
$f(2)=f(1)+\frac{2}{3\cdot 2\cdot1}=\frac{1}{2}=\frac{1}{2}+\frac{1}{3}=\boxe d{\frac{5}{6}}$ (note that $3!=6\\$)
$f(3)=f(2)+\frac{3}{4\cdot3\cdot2\cdot1}=\frac{5}{6 }+\frac{1}{8}=\frac{46}{48}=\boxed{\frac{23}{24}}$ (note that $4!=24$)

Can you guess what has to be shown ?
• May 11th 2008, 07:01 AM
Soroban
Hello, Tangera!

This is quite a task . . .

Quote:

Given that: . $f(n) \:= \:\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} +\hdots + \frac{n}{(n+1)!}$ .where $n$ is a positve integer.

(a) Find the values of $f(1),\:f(2)\text{ and }f(3).$

(b) Make a conjecture for a formula for $f(n)$ in terms of $n$

(c) Prove it using mathematical induction.

$(a) \quad f(1) \:=\: \frac{1}{2}\qquad f(2) \:=\:\frac{5}{6}\qquad f(3) \:=\: \frac{23}{24}$

(b) The denominator seems to be: $(n+1)!$
. . .The numerator seems to be "one less": $(n+1)!-1$

The function seems to be: . $f(n) \;=\;\frac{(n+1)!-1}{(n+1)!}$

$(c)\;\;S(n)\!:\;\;\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \hdots + \frac{n}{(n+1)!} \;=\;\frac{(n+1)!-1}{(n+1)!}$

Verify $S(1)\!:\;\;\frac{1}{2!} \:=\:\frac{2!-1}{2!} \:=\:\frac{1}{2}$ . . . True

Assume $S(k)\!:\;\;\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \hdots + \frac{k}{(k+1)!} \:=\:\frac{(k+1)!-1}{(k+1)!}$

Add $\frac{k+1}{(k+2)!}$ to both sides:

. . $\underbrace{\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \hdots + \frac{k+1}{(k+2)!}} \;=\;\frac{(k+1)!-1}{(k+1)!} +\frac{k+1}{(k+2)!}$

The left side is the left side of $S(k+1).$
We must show that the right side is the right side of $S(k+1)\!:\;\;\frac{(k+2)!-1}{(k+2)!}$

The right side is: . ${\color{blue}\frac{k+2}{k+2}}\!\cdot\!\frac{(k+1)!-1}{(k+1)!} + \frac{k+1}{(k+2)!} \;=\;\frac{(k+2)[(k+1)! - 1] + (k+1)}{(k+2)!}$

. . $= \;\frac{\overbrace{(k+2)(k+1)!}^{\text{This is }(k+2)!}\, \overbrace{- (k+2) + (k+1)}^{\text{This is -1}}}{(k+2)!} \;=\;\frac{(k+2)!-1}{(k+2)!}$ . . . There!

The inductive proof is complete.

• May 15th 2008, 07:28 AM
Tangera
Hello Soroban and flyingsquirrel! Thank you very much for your help! :D