The $n$’th Harmonic number is defined as follows: $$H_n = 1 + \frac12 + \frac13 +\ldots + \frac1n$$ for $n\geq 1$.
a) Show that for all $n≥0$: $H_{2^n} \geq 1+\frac n2$.
b) Show that for all $n≥0$: $H_{2^n} \leq 1+n$.
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See this question. – Winther Mar 12 '15 at 04:51
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1I don't think you have the question right. It looks like Hn=n, so a is easy and b is false. A tutorial for writing math on this site is here As this is your fourth question, you could learn to format things. – Ross Millikan Mar 12 '15 at 04:52
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@Tiina I edited the question with proper math-formatting and made some assumptions about what I thought you meant so please take a look and see if it is correct (click 'Edit' and 'Rollback' to cancel the edit if it is wrong). – Winther Mar 12 '15 at 04:55
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@Winther: That was a huge guess about what OP wanted. Certainly the $H_n$ part is right and I suspect the rest. I would suggest making OP fix things when they are this bad. – Ross Millikan Mar 12 '15 at 04:58
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Homework problem? What have you tried? – Nishant S Mar 12 '15 at 05:00
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it is not a homework question, it is a sample question for my midterm. – Tiina Mar 12 '15 at 05:40
2 Answers
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(a). Base case: for $n = 0,$ we have $H_{2^n}= H_1 = 1,$ and $1 + 0/2 = 1.$
Induction Step: Let's assume $k \geq 0$ is an arbitrary number, and it holds for $k$: $H_{2^k} \geq 1+ k/2.$ Then we prove it for $k+1$: $$\begin{eqnarray} H_{2^{k+1}} & = & H_{2^k} + \underbrace{\frac{1}{2^k+1} + \ldots + \frac{1}{2^{k+1}}}_{\text{($2^k$ terms)}} \\ & \geq &1 + k/2 + \underbrace{\frac{1}{2^k+1} + \ldots + \frac{1}{2^{k+1}}}_{\text{($2^k$ terms)}} \\ & \geq & 1 + k/2 + \underbrace{\frac{1}{2^{k+1}} + \ldots + \frac{1}{2^{k+1}}}_{\text{($2^k$ terms)}} \\ & = & 1 + k/2 + \frac{2^k}{2^{k+1}} \\ & = & 1 + k/2 +1/2 \\ & = & 1+ (k+1)/2 \end{eqnarray}$$ so, it holds for $k+1.$ Therefore, by induction we proved it is true for all $n \geq 0.$
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Dear @Tiina, I've edited your answer, see if it's fine. For future reference here is a link for MathJax: http://meta.matheducators.stackexchange.com/questions/93/mathjax-basic-tutorial-and-quick-reference – Krish Mar 12 '15 at 11:10
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For part b use (make a sketch to see why): $H_n \leq 1 + \int_1^n \frac{dx}{x} = 1 + \log n$. – Winther Mar 12 '15 at 11:15
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@Winther: There is an alternate way to do it. Please see my solution below. – Krish Mar 12 '15 at 11:43
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@Tina I restored parts of your formulas that were deleted in a previous edit, and reworded the start of your induction step: in particular, at the beginning of an inductive step one says something like "it holds for $k$" without specifying which $k$. Also, your ">" inequality was not true when $k=0$ (because $2^0 + 1 = 2^{0+1} = 2$), so I changed it to $\geq$, which makes the inequality true for all $k\geq 0$. Sorry for doing rather than suggesting, but the changes were complicated to describe; you can put back any parts you don't like. – David K Mar 12 '15 at 12:15
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(b) Assume it is true for $k.$ We want to prove that it's true for $k+1.$ First note that, $H_{2^{k+1}} = H_{2^k} +\frac{1}{2^k+1} +….+ \frac{1}{2^{k+1}}.$ Now, for each $k \in \mathbb N, 2^k < 2^k+1, 2^k<2^k+2, \cdots, 2^k<2^{k+1}.$ So $\frac{1}{2^k+1} +….+ \frac{1}{2^{k+1}} \leq 1.$
Krish
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The last sentence has a correct formula, but it's not clear exactly how you derived it from $2^k < 2^{k+1}.$ I think I would replace $2^k < 2^{k+1}$ with a couple of other steps. – David K Mar 12 '15 at 12:25
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@DavidK: There was a typo and some part was missing. I didn't realize. Sorry for that. Now fixed. Please take a look and see if there is any mistake. – Krish Mar 12 '15 at 13:14
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