Question

Prove that \[1+{1 \over 1!}+{1 \over 2!}+{1 \over 3!}+\cdots < 3\]
WITHOUT using the fact that LHS=e.

Answer 1

its a telescoping geometric sequence and need to show that |r| <1


its \[\sum_{n=0}^{\infty} 1/n \times1/(n+1)\]

Answer 2

Doesn't that sum converge to 1?

Answer 3

You can show
\[\frac 1{n!} \le \frac 3{2^n}\\
\sum_{i =0}^\infty
\frac 1{n!} \le \sum_{i =0}^\infty\frac 3{2^n}=3\\
\]

Answer 4

@eliassaab how can you show that\[\sum_{i=0}^{\infty}{1\over n!}\le\sum_{i=0}^{\infty}{3\over 2^n}\]?

Answer 5

and it should be from i=1 I think

Answer 6

Prove by induction that for every n
\[\frac 1{n!} \le \frac 3{2^n}\\

\]

Answer 7

\[\sum_{i=1}^{\infty}{1\over n!}\le\sum_{i=1}^{\infty}{3\over 2^n}=3\]induction you say?
I guess I'll try it...

Answer 8

oh I got my i's and n's mixed up :S
\[\sum_{n=1}^{\infty}{1\over n!}\le\sum_{n=1}^{\infty}{3\over 2^n}=3\]

Answer 9

wait, LHS starts from n=0, RHS starts from n=1, eh?
(if it is to =3)

Answer 10

..and I am having trouble proving this by induction anyway, which is no surprise since I am pretty bad at it :(

Answer 11

\[
\frac 1{1!} \le \frac 3{2^1}\\
\frac 1{2!} \le \frac 3{2^2}\\
\frac 1{3!} \le \frac 3{2^3}\\

\]
Now suppose that for n> 2
\[\frac 1{n!} \le \frac 3{2^n}\\
\]
then
\[
\frac 1{(n+1)!} =\frac 1{n!(n+1)}
\le \frac 3{2^n}\frac 1{(n+1)}\\
\le \frac 3{2^n}\frac 1{2}=\frac 3{2^{n+1}}
\\
\]

Answer 12

I did it for you

Answer 13

Wow. Thanks for the proof. :))

Answer 14

thanks, but what am I right about the LHS starting from n=0 and the RHS starting from n=1 ?
or am I losing it?

Answer 15

...or does that even matter?

Answer 16

@TuringTest Let me fix it for the starting n=0.

Answer 17

I can see your proof is correct, but thank you for the detail :)

Answer 18

You can show that
\[\frac 1 {n!} \le \frac {3}{2^{n+1}}
\]
for
\[
n \ge 4
\]
So
\[\sum_{i =0}^\infty
\frac 1{n!} \le
\sum_{n =0}^3
\frac 1{n!}+
\sum_{n =4}^\infty\frac 3{2^{n+1}}\le 3\\
\]

Answer 19

and that first statement can be shown with induction the exact same way I assume, only for \(n\ge4\), right?