Question

Easy brain teaser:
Evaluate
\[\sum_{k=0}^n \frac{k}{(k+1)!}\]

Answer 1

\[=0+\frac12+\frac26+\frac3{24}+\frac4{120}+\cdots\]\[=0+\frac12+\frac13+\frac18+\frac1{30}+\cdots\]

Answer 2

\[\approx0+0.5+0.333+0.125+0.033+\cdots\]\[=0.5+0.457+0.033\]\[=0.5+0.490\]\[=0.99\]\[\approx1\]

Answer 3

It's not until infinity. :P

Answer 4

you add consecutive terms until you get + n/(n+1)!
so it's 0 + ... + n/(n+1)! the number of terms you want to add depends on your preference.

Answer 5

k=k+1-1 in num can help you.

Answer 6

I want the answer to not have a summation. (Of course, I know the answer.)

Answer 7

sig(1/k!-1/k+1!)=1-0=1

Answer 8

The answer I wanted was \(1-\frac{1}{(n+1)!}\) but this will do.

Answer 9

n is going to inf then i/n+1! going to 0

Answer 10

ok i thought uper was inf

Answer 11

yeap 1-1/n+1! is answer

Answer 12

dont forget k=k+1-1