Proof by induction (how did I do?)

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• May 14th 2011, 01:04 PM
ElectroSpecter
Proof by induction (how did I do?)
Hi everyone, I'm new to the forums. I'm also fairly new to proofs, and I still commonly make mistakes, specifically regarding quantification errors and circularity. However, I believe that I have gotten this particular proof nailed down pretty well. I was wondering if I could get some feedback on whether or not what I did seems correct. I'm wondering specifically about the fourth line of the induction part, where I said "Note that this equation may also be written as [blah]." Should I elaborate on why, or is it obvious enough?

(Also, sorry about the alignment of some of the longer equations, I can't seem to get it to work like I normally do).

Any help is greatly appreciated!

Theorem: For all $\displaystyle n \in \mathbb{N}$, $\displaystyle \frac{1}{1 \cdot 2}+ \frac{1}{2 \cdot 3}+ \cdot \cdot \cdot + \frac{1}{n(n+1)}=\frac{n}{n+1}.$

Proof: We will proceed by induction.

i) Base step: Let $\displaystyle n = 1$. We see that $\displaystyle \displaystyle \frac{1}{1 \cdot 2} = \frac{1}{2}$. We also see that
$\displaystyle \frac{n}{n+1} & = \frac{1}{1+1}$
$\displaystyle = \frac{1}{2}.$
Therefore, $\displaystyle P_1$ is true.

ii) Inductive step: Let $\displaystyle k \in \mathbb{N}$. Assume that $\displaystyle \displaystyle \frac{1}{1 \cdot 2}+ \frac{1}{2 \cdot 3}+ \cdot \cdot \cdot + \frac{1}{k(k+1)}=\frac{k}{k+1}$. We would like to show that $\displaystyle \displaystyle \frac{1}{1 \cdot 2}+ \frac{1}{2 \cdot 3}+ \cdot \cdot \cdot + \frac{1}{(k+1)(k+2)}=\frac{k+1}{k+2}$. Note that this equation may also be written as $\displaystyle \displaystyle \frac{1}{1 \cdot 2}+ \frac{1}{2 \cdot 3}+ \cdot \cdot \cdot + \frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)}=\frac{k+1}{k+2}$. By our assumption, we see that
$\displaystyle \displaystyle \frac{1}{1 \cdot 2}+ \frac{1}{2 \cdot 3}+ \cdot \cdot \cdot + \frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)} & = \frac{k}{k+1} + \frac{1}{(k+1)(k+2)}$
$\displaystyle & = \frac{k}{k+1}\Bigg(\frac{k+2}{k+2}\Bigg) + \frac{1}{(k+1)(k+2)}$
$\displaystyle = \frac{k^2+2k+1}{(k+1)(k+2)}$
$\displaystyle = \frac{(k+1)^2}{(k+1)(k+2)}$
$\displaystyle = \frac{k+1}{k+2}.$
Therefore, $\displaystyle P_{k+1}$ is true.

Furthermore, by the PMI, $\displaystyle P_n$ is true for all $\displaystyle n \in \mathbb{N}$.
• May 14th 2011, 03:24 PM
Archie Meade
That's good.
• May 14th 2011, 03:25 PM
emakarov
Your proof is correct. Concerning the line with "Note that this equation may also be written as," you did not apply any algebraic transformation, you just wrote another term in the sum that was implicit in the previous equation. I would write,

"We would like to show that $\displaystyle \displaystyle \frac{1}{1 \cdot 2}+ \frac{1}{2 \cdot 3}+ \cdot \cdot \cdot + \frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)}=\frac{k+1}{k+2}$"

and omit the following sentence.