Approximate value of ##E=1/2! +1/3!+1/4!+... ##

[littlemathquark]

Homework Statement

$n\in \mathbb{Z^+}$ and given $1/n! > 1/(n+1)! +1/(n+2)!+...$
find the approximate value of $E=1/2! +1/3!+1/4!+...$

Relevant Equations

$n\in \mathbb{Z^+}$ and if we know $1/n! > 1/(n+1)! +1/(n+2)!+...$
find the approximate value of $E=1/2! +1/3!+1/4!+...$

I know $1+1/1!+1/2!+1/3!+...=e=2,71..$ but I have to find approximate value of E using inequality.

 

[berkeman]

but I have to find

Is this for schoolwork?

 

[littlemathquark]

Yes, it's in a math test.

 

[berkeman]

Then you need to show more effort. We do not take your math tests for you.

 

[littlemathquark]

The following are the test choices
0,5 0,6 0,7 0,8 0,9

 

[littlemathquark]

Then you need to show more effort. We do not take your math tests for you.

I know but I need an idea.

 

[fresh_42]

I know but I need an idea.

Idea: Bohr-Mollerup uses the convexity of the logarithm of $\Gamma(x).$

 

[littlemathquark]

Thank you but I dont know that theorem but I found this idea
İt's clear E>0,5

E<1/2! +1/2!
E<(1/2! +1/3!)+1/3!
E<(1/2!+1/3! +1/4!)+1/4!
E<(1/2!+1/3! +1/4!+1/5!)+1/5!

 

[littlemathquark]

E<(1/2!) +1/2!=1 and added number 1/2!=0,5
E<(1/2! +1/3!)+1/3!=5/6=0,83 and added number 1/3!=0,16
E<(1/2!+1/3! +1/4!)+1/4!=3/4=0,75 and added number 1/4!=0,04

It can be said that after this stage, since the numbers added will be very small, the number E approaches 0.7 by decreasing.

 

[littlemathquark]

Well, $1/2^n$ is a lower bound for $n \geq 1$, though hardly a tight one. Edit: Thus $1+1/2+...+1/2^n+...=2*$ would be a lower bound, albeit obviously not too tight.

*Converging to, as n grows.

İs $1/2^n$ is a lower bound or upper bound?

 

[WWGD]

İs $1/2^n$ is a lower bound or upper bound?

Edit: My bad, I realize I misread; let me delete it. This is my idea: As a loweer bound, add the first 10 terms $1+(1/2!)+...+(1/10!)$ and then use a geometric progression starting with $(1/11!)$ and common ratio $(1/11)$,

Then we have the bound $1+(1/2!)+....+(1/10!)+ (1/39916800)[\frac{1}{1-(1/11)}$.]

 

[WWGD]

My bad, I realize I misread; let me delete it. This is my idea: As an upper bound, add the first 10 terms $1+(1/2!)+...+(1/10!)$ and then use a geometric progression starting with $(1/11!)$ and common ratio $(1/11)$,

Then we have the bound $1+(1/2!)+....+(1/10!)+ (1/39916800)[\frac{1}{1-(1/11)}$.]

 

[mathwonk]

the given inequality is an upper bound for the error in approximating E by its first n-1 terms.

Unfortunately, use of a small hand held calculator also introduced, for me, roundoff error. My computation of the first 12 terms gave me .718281829, which should have been an underestimate, with error less than 1/13!, but is obviously an overestimate in the last digit.