Proof by induction that $\sum_{i=1}^n \frac{i}{(i+1)!}=1- \frac{1}{(n+1)!}$ Prove via induction that $\sum_{i=1}^n \frac{i}{(i+1)!}=1- \frac{1}{(n+1)!}$
Having a very difficult time with this proof, have done pages of work but I keep ending up with 1/(k+2).  Not sure when to apply the induction hypothesis and how to get the result $1- \frac{1}{(n+2)!}$.  Please help!
thanks guys, youre the greatest!
 A: Hint: $$1-\frac1{(n+1)!}+\frac{n+1}{(n+2)!} = 1-\frac{n+2}{(n+2)!}+\frac{n+1}{(n+2)!} = 1-\frac{1}{(n+2)!}.$$
A: Induction assumption assumes your statement holds for $n=k$
Step after induction assumption. Let  $n=k+1$ then the right hand side of your statement is,
$$ 1-\cfrac{1}{(k+1)!}+\cfrac{k+1}{(k+2)!}.$$
Take common denominator of last two terms and you get
$$ 1+\cfrac{-k+-2+k+1}{(K+2)!}=1-\cfrac{1}{(k+2)!}$$
A: $$\frac{n}{(n+1)!}=\frac{n+1-1}{(n+1)!}=\frac{n+1}{(n+1)!}-\frac{1}{(n+1)!}=\frac{1}{n!}-\frac{1}{(n+1)!}$$
Note that this idea can also be used to make the sum into a telescopic sum.
A: You can also prove this by power series manipulation (generating functions).  Note first that changing the sum by starting at $i=0$ instead of $i=1$ doesn't change its value. Compute as follows:
\begin{align}\sum_{n=0}^\infty \sum_{i=1}^n \frac{i}{(i+1)!}x^{n+1} &= \sum_{n=0}^\infty \sum_{i=0}^n \frac{i}{(i+1)!}x^{n+1}
\\\\&= \sum_{i=0}^\infty \sum_{n=i}^\infty \frac{i}{(i+1)!}x^{n+1}
\\\\ &= \sum_{i=0}^\infty \frac{i}{(i+1)!} \left( \sum_{n=-1}^\infty x^{n+1} - \sum_{n=-1}^{i-1} x^{n+1}\right)
\\\\&= \sum_{i=0}^\infty \frac{i}{(i+1)!} \left(\frac{1}{1-x}-\frac{1-x^{i+1}}{1-x} \right)
\\\\&= \sum_{i=0}^\infty \frac{i}{(i+1)!}\frac{x^{i+1}}{1-x}
\\\\&= \frac{x^2}{1-x} \sum_{i=0}^\infty \frac{ix^{i-1}}{(i+1)!}
\\\\&= \frac{x^2}{1-x} \frac{d}{dx} \sum_{i=0}^\infty \frac{x^i}{(i+1)!}
\\\\&= \frac{x^2}{1-x} \frac{d}{dx} \frac{e^x-1}{x}
\\\\&= \frac{x^2}{1-x}  \frac{xe^x-(e^x-1)}{x^2}
\\\\&=\frac{1}{1-x}-e^x
\\\\&=\sum_{n=0}^\infty x^n - \sum_{n=0}^\infty \frac{x^n}{n!}
\\\\&= \sum_{n=0}^\infty \left(1-\frac{1}{n!}\right)x^n
\\\\&= \sum_{n=1}^\infty \left(1-\frac{1}{n!}\right)x^n
\\\\&= \sum_{n=0}^\infty \left(1-\frac{1}{(n+1)!}\right)x^{n+1} \end{align}
Coefficients of like powers of $x$ in two equal power series must be equal, proving that $$\sum_{i=1}^n \frac{i}{(i+1)!} = 1-\frac{1}{(n+1)!},$$ as desired.
This can all be justified either as formal power series manipulation or as calculations with absolutely convergent series for $|x| < 1.$
