Proof by induction of a problem

[marcin w]

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 $$\sum_{k = m}^{m + n}{a}_{k}$$
I first define:
$$\sum_{k = m}^{m}{a}_{k} = {a }_{ m}$$
Then assuming I have defined $$\sum_{k = m}^{n}{a}_{k}$$ for a fixed n >= m, I further defined:
$$\sum_{ k=m}^{n+1 } {a }_{k } =( \sum_{k=m }^{ n} { a}_{ k} )+ { a}_{n+1 }$$

Part 2 requires me to prove by induction that for all n >= 1, we have the assertion (call it A(n)):
$$\sum_{k=n+1 }^{ 2n} \frac{1 }{k } = \sum_{ m=1}^{ 2n} \frac{ {(-1) }^{m+1 } }{m }$$

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:
$$\frac{ 1}{ k+1} + \frac{1 }{k+2 } +...+ \frac{ 1}{ 2k} = 1 - \frac{ 1}{2 } + \frac{1 }{3 } - \frac{ 1}{ 4} +...+ \frac{ {(-1) }^{2k+1 } }{ 2k}$$

where the last term on the RHS simplifies to $$- \frac{ 1}{2k }$$.
I have to show that A(k+1) is true:
(*)$$\frac{ 1}{ k+2} + \frac{1 }{k+3 } +...+ \frac{ 1}{ 2(k+1)} = 1 - \frac{ 1}{2 } + \frac{1 }{3 } - \frac{ 1}{ 4} +...+ \frac{ {(-1) }^{2(k+1)+1 } }{ 2(k+1)}$$

where the last term on the RHS simplifies to $$- \frac{1 }{ 2(k+1)}$$
Starting with A(k), I add $$\frac{1 }{ 2k+1} + \frac{ 1}{ 2k+2} - \frac{ 1}{ k+1}$$ to each side and obtain:

$$\frac{ 1}{k+1 } + \frac{ 1}{k+2 } +...+ \frac{ 1}{2k } - \frac{1 }{k+1 } + \frac{1 }{ 2k+1} + \frac{1 }{2k+2 } =1- \frac{ 1}{ 2} + \frac{ 1}{ 3} - \frac{ 1}{ 4} +...- \frac{1 }{2k+2 } - \frac{1 }{ k+1} + \frac{ 1}{ 2k+1} + \frac{1 }{2k+2 }$$
After subtracting out the two like terms on each side, the left hand side becomes the sum
$$\sum_{ k=(n+1)+1}^{ 2(n+1)} \frac{1 }{k }$$


and the terms $$\frac{ 1}{ 2k+2}- \frac{ 1}{ k+1}$$ on the right simplify to $$\frac{ -1}{2(k+1) }$$, so the right side resembles the right side in (*). Therefore, the RHS becomes the sum $$\sum_{ m=1}^{2(n+1) } \frac{ {(-1) }^{m+1} }{2m }$$

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.

 

[HallsofIvy]

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 $$\sum_{k = m}^{m + n}{a}_{k}$$
I first define:
$$\sum_{k = m}^{m}{a}_{k} = {a }_{ m}$$
Then assuming I have defined $$\sum_{k = m}^{n}{a}_{k}$$ for a fixed n >= m, I further defined:
$$\sum_{ k=m}^{n+1 } {a }_{k } =( \sum_{k=m }^{ n} { a}_{ k} )+ { a}_{n+1 }$$

Part 2 requires me to prove by induction that for all n >= 1, we have the assertion (call it A(n)):
$$\sum_{k=n+1 }^{ 2n} \frac{1 }{k } = \sum_{ m=1}^{ 2n} \frac{ {(-1) }^{m+1 } }{m }$$

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.

Your teacher might not appreciate that! Saying something is true is not proving that it is true.

Now, assuming the assertion is true for some k:
$$\frac{ 1}{ k+1} + \frac{1 }{k+2 } +...+ \frac{ 1}{ 2k} = 1 - \frac{ 1}{2 } + \frac{1 }{3 } - \frac{ 1}{ 4} +...+ \frac{ {(-1) }^{2k+1 } }{ 2k}$$

where the last term on the RHS simplifies to $$- \frac{ 1}{2k }$$.
I have to show that A(k+1) is true:
(*)$$\frac{ 1}{ k+2} + \frac{1 }{k+3 } +...+ \frac{ 1}{ 2(k+1)} = 1 - \frac{ 1}{2 } + \frac{1 }{3 } - \frac{ 1}{ 4} +...+ \frac{ {(-1) }^{2(k+1)+1 } }{ 2(k+1)}$$

where the last term on the RHS simplifies to $$- \frac{1 }{ 2(k+1)}$$
Starting with A(k), I add $$\frac{1 }{ 2k+1} + \frac{ 1}{ 2k+2} - \frac{ 1}{ k+1}$$ to each side and obtain:

$$\frac{ 1}{k+1 } + \frac{ 1}{k+2 } +...+ \frac{ 1}{2k } - \frac{1 }{k+1 } + \frac{1 }{ 2k+1} + \frac{1 }{2k+2 } =1- \frac{ 1}{ 2} + \frac{ 1}{ 3} - \frac{ 1}{ 4} +...- \frac{1 }{2k+2 } - \frac{1 }{ k+1} + \frac{ 1}{ 2k+1} + \frac{1 }{2k+2 }$$
After subtracting out the two like terms on each side, the left hand side becomes the sum
$$\sum_{ k=(n+1)+1}^{ 2(n+1)} \frac{1 }{k }$$


and the terms $$\frac{ 1}{ 2k+2}- \frac{ 1}{ k+1}$$ on the right simplify to $$\frac{ -1}{2(k+1) }$$, so the right side resembles the right side in (*). Therefore, the RHS becomes the sum $$\sum_{ m=1}^{2(n+1) } \frac{ {(-1) }^{m+1} }{2m }$$

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.

In going form n to n+1, two things happen. First, since the sum starts at n+1 in the first case and (n+1)-1 in the second, you are missing the first term. Second, since the sum ends at 2n in the first case and 2(n+1)= 2n+ 2 in the second, you will have two new terms, 2n+1 and 2n+ 2. That is,
$$\sum_{k= (n+1)+1}^{2(n+1)} \frac{1}{k}= \sum_{k= n+1}^{2n} \frac{1}{k}+ \frac{1}{2n+1}+ \frac{1}{2n+2}- \frac{1}{n+1}$$
Now, replace that sum by
$$\sum_{m=1}^{2n} \frac{(-1)^m+1}{m}$$
and do the algebra.

You might want to use the fact that
$$\frac{1}{2n+2}- \frac{1}{n+1}= \frac{1- 2}{2n+2}= -\frac{1}{2n+2}$$

 

[marcin w]

Thanks, your answer was helpful and much appreciated.