Mathematical Induction (2)

**Tangera** · May 11th 2008, 05:48 AM

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.

Thank you in advance!

**flyingsquirrel** · May 11th 2008, 06:21 AM

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 ?

**Soroban** · May 11th 2008, 07:01 AM

Hello, Tangera!

This is quite a task . . .

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.

**Tangera** · May 15th 2008, 07:28 AM

Hello Soroban and flyingsquirrel! Thank you very much for your help!

Thread: Mathematical Induction (2)

LinkBack

Thread Tools

Display

Mathematical Induction (2)

Similar Math Help Forum Discussions

Mathematical Induction

Mathematical induction

Rigorously prove the principle of complete induction from mathematical induction??

mathematical induction

Mathematical Induction

Search Tags