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Proof by induction of a problem 
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#1
Feb2507, 04:05 PM

P: 12

I'm working out a problem that requires me to proof a result by induction. I have worked out what I think is a correct proof, but I would like for somebody to look over it and give me feedback.
Part 1: Give a reasonable definition of the symbol [tex]\sum_{k = m}^{m + n}{a}_{k}[/tex] I first define: [tex] \sum_{k = m}^{m}{a}_{k} = {a }_{ m} [/tex] Then assuming I have defined [tex]\sum_{k = m}^{n}{a}_{k}[/tex] for a fixed n >= m, I further defined: [tex] \sum_{ k=m}^{n+1 } {a }_{k } =( \sum_{k=m }^{ n} { a}_{ k} )+ { a}_{n+1 } [/tex] Part 2 requires me to prove by induction that for all n >= 1, we have the assertion (call it A(n)): [tex]\sum_{k=n+1 }^{ 2n} \frac{1 }{k } = \sum_{ m=1}^{ 2n} \frac{ {(1) }^{m+1 } }{m } [/tex] I approach this problem as I would have any proof by induction. The base case A(1) is true so I won't write it out here. Now, assuming the assertion is true for some k: [tex] \frac{ 1}{ k+1} + \frac{1 }{k+2 } +...+ \frac{ 1}{ 2k} = 1  \frac{ 1}{2 } + \frac{1 }{3 }  \frac{ 1}{ 4} +...+ \frac{ {(1) }^{2k+1 } }{ 2k} [/tex] where the last term on the RHS simplifies to [tex] \frac{ 1}{2k } [/tex]. I have to show that A(k+1) is true: (*)[tex]\frac{ 1}{ k+2} + \frac{1 }{k+3 } +...+ \frac{ 1}{ 2(k+1)} = 1  \frac{ 1}{2 } + \frac{1 }{3 }  \frac{ 1}{ 4} +...+ \frac{ {(1) }^{2(k+1)+1 } }{ 2(k+1)} [/tex] where the last term on the RHS simplifies to [tex] \frac{1 }{ 2(k+1)} [/tex] Starting with A(k), I add [tex] \frac{1 }{ 2k+1} + \frac{ 1}{ 2k+2}  \frac{ 1}{ k+1} [/tex] to each side and obtain: [tex]\frac{ 1}{k+1 } + \frac{ 1}{k+2 } +...+ \frac{ 1}{2k }  \frac{1 }{k+1 } + \frac{1 }{ 2k+1} + \frac{1 }{2k+2 } =1 \frac{ 1}{ 2} + \frac{ 1}{ 3}  \frac{ 1}{ 4} +... \frac{1 }{2k+2 }  \frac{1 }{ k+1} + \frac{ 1}{ 2k+1} + \frac{1 }{2k+2 } [/tex] After subtracting out the two like terms on each side, the left hand side becomes the sum [tex] \sum_{ k=(n+1)+1}^{ 2(n+1)} \frac{1 }{k } [/tex] and the terms [tex] \frac{ 1}{ 2k+2} \frac{ 1}{ k+1} [/tex] on the right simplify to [tex] \frac{ 1}{2(k+1) } [/tex], so the right side resembles the right side in (*). Therefore, the RHS becomes the sum [tex] \sum_{ m=1}^{2(n+1) } \frac{ {(1) }^{m+1} }{2m } [/tex] I'm wasn't sure if I was supposed to use the definition of part 1 in part 2 and I didn't, but I'm wondering if that would have made things easier. Thanks. 


#2
Feb2607, 07:28 AM

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PF Gold
P: 39,682

[tex]\sum_{k= (n+1)+1}^{2(n+1)} \frac{1}{k}= \sum_{k= n+1}^{2n} \frac{1}{k}+ \frac{1}{2n+1}+ \frac{1}{2n+2} \frac{1}{n+1}[/tex] Now, replace that sum by [tex]\sum_{m=1}^{2n} \frac{(1)^m+1}{m}[/tex] and do the algebra. You might want to use the fact that [tex]\frac{1}{2n+2} \frac{1}{n+1}= \frac{1 2}{2n+2}= \frac{1}{2n+2}[/tex] 


#3
Feb2607, 08:52 PM

P: 12

Thanks, your answer was helpful and much appreciated.



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