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

MATH220 December 2010

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Question 06 (a)
Use mathematical induction to prove the following. Prove that ${\frac {1}{2!}}+{\frac {2}{3!}}+\dots +{\frac {n}{(n+1)!}}=1-{\frac {1}{(n+1)!}}$ is true for all $n\in \mathbb {N}$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Recall the ingredients for a proof by induction: 1. Prove a base case. 2. Assume the claim is true for some $k$ 3. Prove the claim is true for $k+1$

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 $n\in \mathbb {N}$.

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