Proving "1+1+1/2!+1/3!+...+1/n! < 3

[topengonzo]

Homework Statement

Prove that 1+1+1/2!+1/3!+...+1/n! < 3

Homework Equations

None

The Attempt at a Solution

Any suggestion how can I start?
I don't want the solution that 1+1+1/2!+1/3!+...+1/n!=(1+1/n)^n and lim (1+1/n)^n = e<3 since the 2nd part is to prove (1+1/n)^n <3 . I don't want a solution. I just want to know how to start with it.

 

[tiny-tim]

welcome to pf!

hi topengonzo! welcome to pf!

hint: can you see a way to prove that 1/2!+1/3!+...+1/n! < 1 ?

 

[topengonzo]

hi tiny-tim

this is the first thing i thought of and i actually worked without the first 2 terms. Prove by induction on this problem is impossible i think. So i don't have a clue how to solve it. :S

 

[tiny-tim]

1/2+1/2.3+1/2.3.4+...+ < ?

 

[sunjin09]

Homework Statement

Prove that 1+1+1/2!+1/3!+...+1/n! < 3

Homework Equations

None

The Attempt at a Solution

Any suggestion how can I start?
I don't want the solution that 1+1+1/2!+1/3!+...+1/n!=(1+1/n)^n and lim (1+1/n)^n = e<3 since the 2nd part is to prove (1+1/n)^n <3 . I don't want a solution. I just want to know how to start with it.

hint: does the original series look familiar to you at all? Have you compared it to any series you already know?
(If you haven't learned taylor expansion ignore what I said here)

 

[topengonzo]

i don't get what u mean. I am thinking of proving that sum of next terms is less than equal to current term that is:
(1/3!) + (1/4!)+(1/5!)+...(1/infinity!)<(1/2!)
(1/4!) + (1/5!)+(1/6!)+...(1/infinity!)<(1/3!)
... (1/(n+1)!) + (1/(n+2)!)+... (1/infinity!)<(1/n!)

Should I reach a solution with this method or can I solve it this way?

 

[topengonzo]

hint: does the original series look familiar to you at all? Have you compared it to any series you already know?
(If you haven't learned taylor expansion ignore what I said here)

I think its taylor expansion will give me e<3 and solved. Is there another way to solve it?

 

[sunjin09]

deleted

 

[tiny-tim]

sunjin09, please don't give the full answer

 

[topengonzo]

what about 1+1+1/2+1/4+1/8+... compare to your series? what's the sum?

THANK YOU VERY VERY MUCH!

So I prove 1/k! < 1/ 2^(k-1) for k>=3 by induction and then i can say my series < 1+ 1 + 1/2 + 1/4 + 1/8 + ... which is geometric series with r=1/2 and first term 1
implies 1+ 1/(1-1/2) = 3

 

[checkitagain]

1/2+1/2.3+1/2.3.4+...+ < ?

That is the same as \dfrac{1}{2} \ + \ \dfrac{1}{2}\cdot 3 \ + \ \dfrac{1}{2}\cdot3\cdot4 \ \ + \ ... \ + \ &lt; \ ?



(You're using the dots for multiplication. Written out horizontally,
grouping symbols are needed.)



Instead, it could be correctly shown horizontally as:


1/2 + 1/(2*3) + 1/(2*3*4) + ... + < ?


or as



1/2 \ + \ 1/(2\cdot 3) \ + \ 1/(2\cdot3\cdot4) \ + \ ... \ + \ &lt; \ ?