and 
n
.
Prove that
Let 
Prove that
. Prove that
are nonzero real numbers, find the minimum value of the expression![\[
P = \left| \frac{b + c + d}{a} \right| + \left| \frac{c + d + a}{b} \right| + \left| \frac{d + a + b}{c} \right| + \left| \frac{a + b + c}{d} \right|.
\]](http://latex.artofproblemsolving.com/e/3/e/e3e63937c4badb9bcb47e30778f2165e2df5ae46.png)
Y by
Is there a way to prove
\frac{1}{(1+1)!}+\frac{2}{(2+1)!}+...+\frac{n}{(n+1)!}=1-\frac{1}{{n+1)!}
without induction and using only combinatorial arguments?
Induction proof wasn't quite as pleasing for me.
\frac{1}{(1+1)!}+\frac{2}{(2+1)!}+...+\frac{n}{(n+1)!}=1-\frac{1}{{n+1)!}
without induction and using only combinatorial arguments?
Induction proof wasn't quite as pleasing for me.
This post has been edited 1 time. Last edited by MathBot101101, Apr 21, 2025, 6:45 AM
Reason: im dum
Reason: im dum
![\[ \sum\limits_{i=1}^n \dfrac{i}{(i + 1)!} = 1 - \dfrac{1}{(n+1)!}. \]](http://latex.artofproblemsolving.com/2/8/8/288b06247fd042559aaa1e0c0c101b5b2eba7821.png)
This is indeed a simple induction exercise. As for combinatorial arguments, maybe we can rearrange this into ![\[ \frac{1}{(n+1)!} + \sum\limits_{i=1}^{n} \frac{i}{(i+1)!} = 1,\]](http://latex.artofproblemsolving.com/f/1/a/f1a775f27856ac90581ef10e3fb4f6faaee58a71.png)
but then I'm not sure how to get a probability that is clearly equal to the LHS and is also equal to 
.
)
gives that
(This is induction at heart.)
