Science:Math Exam Resources/Courses/MATH220/December 2010/Question 06 (a)/Solution 1

For $n\in \mathbb {N}$, let $\displaystyle S(n)$ be the claim that

${\frac {1}{2!}}+{\frac {2}{3!}}+\dots +{\frac {n}{(n+1)!}}=1-{\frac {1}{(n+1)!}}$

We prove that $\displaystyle S(n)$ is true for all natural numbers by using induction. First, we prove that $\displaystyle S(1)$ is true which is clear since

$1-{\frac {1}{(1+1)!}}=1-{\frac {1}{2}}={\frac {1}{2!}}$

Now, we assume that $\displaystyle S(k)$ is true for some $k\in \mathbb {N}$ (the induction hypothesis) and prove that $\displaystyle S(k+1)$ is true. That is, we want to show that

${\frac {1}{2!}}+{\frac {2}{3!}}+\dots +{\frac {k}{(k+1)!}}+{\frac {k+1}{((k+1)+1)!}}=1-{\frac {1}{((k+1)+1)!}}$

Using the induction hypothesis, we have

${\begin{aligned}\left({\frac {1}{2!}}+{\frac {2}{3!}}+\dots +{\frac {k}{(k+1)!}}\right)+{\frac {k+1}{((k+1)+1)!}}&=\left(1-{\frac {1}{(k+1)!}}\right)+{\frac {k+1}{(k+2)!}}\\&=1-{\frac {k+2}{(k+2)!}}+{\frac {k+1}{(k+2)!}}\\&=1+{\frac {-k-2+k+1}{(k+2)!}}\\&=1+{\frac {-1}{(k+2)!}}\\&=1-{\frac {1}{((k+1)+1)!}}\end{aligned}}$

Thus $\displaystyle S(k)\Rightarrow S(k+1)$. Hence, by mathematical induction, we have that $\displaystyle S(n)$ is true for all $n\in \mathbb {N}$.