Proof by induction that $\sum_{i=1}^n \frac{i}{(i+1)!}=1- \frac{1}{(n+1)!}$

Question

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!

Answer 1

Hint: $$1-\frac1{(n+1)!}+\frac{n+1}{(n+2)!} = 1-\frac{n+2}{(n+2)!}+\frac{n+1}{(n+2)!} = 1-\frac{1}{(n+2)!}.$$

Answer 2

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 $$ 1+\cfrac{-k+-2+k+1}{(K+2)!}=1-\cfrac{1}{(k+2)!}$$

Answer 3

$$\frac{n}{(n+1)!}=\frac{n+1-1}{(n+1)!}=\frac{n+1}{(n+1)!}-\frac{1}{(n+1)!}=\frac{1}{n!}-\frac{1}{(n+1)!}$$

Note that this idea can also be used to make the sum into a telescopic sum.