To infinity and beyond...

Calculus Level 2

What is the value of ln ( 1 + 1 + 1 2 + 1 6 + 1 24 + 1 120 + . . . ) ? \text{ln }(1+1+\frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \frac{1}{120} +...) \ ?

1 e ~2.718 π 2 6 \frac{\pi^2}{6} 0 ∞ \infty 1 e \frac{1}{e} -1 π \pi

Curtis Clement by Curtis Clement , 23, United Kingdom

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Curtis Clement
Feb 14, 2015

The sequence is equivalent to L n ( ∑ k = 0 ∞ 1 k ! ) = L n ( e ) = 1 Ln(\displaystyle \sum_{k=0}^{\infty}\dfrac{1}{k!}) = Ln(e) = \boxed{1} from the taylor series expansion of e x e^{x} with x {x} = 1: e x = ∑ k = 0 ∞ x k k ! e^x = \displaystyle \sum_{k=0}^{\infty}\dfrac{x^k}{k!}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Crap, I didn't see the natural log function. I'm disappointed in myself.

tytan le nguyen - 6 years, 3 months ago

Log in to reply

Don't worry about it - It didn't know at your age. Maybe I'll post some similar questions to give you an opportunity to redeem yourself ;)

Curtis Clement - 6 years, 3 months ago

Log in to reply

Well, this is a good problem. Keep it up!

tytan le nguyen - 6 years, 3 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...