Question

Prove by the principal of mathematical Induction:
\( \frac{1} {2!} + \frac{2} {3!} +...+ \frac{n} {(n+1)!} = 1 - \frac{1} {(n+1)!} \)

Answer 1

If you type in "huge" or "large" in the equation editor, it'll make all that bigger fwiw. :D

So how would you start off?

Answer 2

The first step wld be to verify that this is true witthe lowest value of n which is 1 in thsi case

Answer 3

Right, our Basis case right?

Answer 4

yup

Answer 5

which obv is true

Answer 6

Next we need to start our IH (induction Hypothesis).
One thing that I always do is the following:
Let P(n) be the statement "insert huge statement here"

Answer 7

then P(n+1) is true

Answer 8

Then i follow in the IH like so:
Spse P(n) is try, show that P(n+1) is true for any arbitrary n.

Answer 9

ya so that is my difficulty

Answer 10

Since the LHS is in a series, when we add "one" we add a whole new term. therefore we would add an additional:
\[\huge + {n+1 \over (n+2)!} = 1 - {1 \over (n+1)!}\]

Answer 11

so on my LHS just imagine that there is everything you wrote from the LHS before my "plus" sign. Does that make sense?

Answer 12

welll y didnt u add a +1 on the right hand side?

Answer 13

Oh whoops... you're right. Good catch.

Answer 14

But since we're suposing that the Basis was true, P(n), we can replace the whole series representation with:
\[\huge 1-{1 \over (n+1)!}\]
and can then combine everything together now (correctly)
\[\huge 1 - {1 \over (n+1)!} + {n+1 \over (n+2)!} = 1 - {1 \over (n+2)!}\]

Answer 15

Think you can take it from there?

Answer 16

idk i didnt follow at all

Answer 17

Let p(n) be the statement \(\frac{1}{2!} + \frac{2}{3!} + ... + \frac{n}{(n+1)!} = 1-\frac{1}{(n+1)!}\)
Consider p(1)
LS = \( \frac{1}{(1+1)!} = \frac{1}{2}\)
RS = \(1-\frac{1}{(1+1)!} = \frac{1}{2}\)
So, P(1) is true

Suppose p(k) is true for some positive integers k.
LS of P(k+1)
= \(\frac{1}{2!} + \frac{2}{3!} + ... + \frac{k}{(k+1)!} +\frac{k+1}{[(k+1)+1]!}\)
= \( 1-\frac{1}{(k+1)!}+\frac{k+1}{(k+2)!}\)
= \( 1-\frac{(k+2) - (k+1)}{(k+2)!}\)
= \(1 -\frac{1}{(k+2)!}\)
= RS of P(k+1)

I think you can continue from here?

Answer 18

I didnt follow one point. The third line where you everything under (k+2)!
How did u get (k+2)-(K+1). I just suck at sequences like I never really learnt it

Answer 19

(k+2)! = (k+2)(k+1)!
\(1-\frac{(1}{(k+1)!}+\frac{(k+1)}{(k+2)!} \)
\(=1-\frac{1(k+2)}{(k+2)(k+1)!}+\frac{(k+1)}{(k+2)!}\)
\(=1-\frac{(k+2)}{(k+2)!}+\frac{(k+1)}{(k+2)!}\)
\(=1-\frac{(k+2)-(k+1)}{(k+2)!}\)

Answer 20

ohhhhhhh i seee lol

Answer 21

hahahahaha

Answer 22

Ok the rest of the proof was really clear. THANKKKKKSSSSS

Answer 23

THAT WAS SERIOUSLY AWESOME

Answer 24

THANKS :))))))))))))))))

Answer 25

Welcome, and thanks! :)