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The question is prove that for every integer greater than or equal to 2

$$\frac{1}{n+1} + \frac{1}{n+2} + \ldots + \frac{1}{2n} \geq \frac{7}{12}$$

So far I have

Base case let $p(2)$ \begin{align*} \frac{1}{2}+ \frac{1}{2+2} + \frac{1}{4} & \geq \frac{7}{12}\\ \frac{5}{6} & \geq \frac{7}{12} \end{align*}

therefore base case is true. Let's assume it is a true statement for $p(n+1)$ for all $n$ greater than or equal to $2.$

\begin{align*} \frac{1}{(n+1)+1} + \frac{1}{(n+1)+2} + \frac{1}{2(n+1)} \geq \frac{7}{12}\\ \frac{1}{n+2} + \frac{1}{n+3} + \frac{1}{2n+2} \geq \frac{7}{12} \end{align*}

then I wrote that if $n$ is greater than or equal to $2$ is a true statement then the inequality is true. This is a proof by induction and I'm not sure if I concluded properly

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    $\begingroup$ Please see this tutorial for proper typesetting. It will help everyone better understand your question. $\endgroup$ – Daniel W. Farlow Feb 18 '15 at 23:51
  • $\begingroup$ Oh perfect thank you very much $\endgroup$ – user212401 Feb 18 '15 at 23:56
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    $\begingroup$ The base case is for $n=2$ is actually $1/3+1/4=7/12$ $\endgroup$ – Henry Feb 19 '15 at 0:01
  • $\begingroup$ 1/3+1/4+1/4 =5/6 $\endgroup$ – user212401 Feb 19 '15 at 0:06
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    $\begingroup$ @user212401 No, there are only two terms in the $n=2$ case. The ellipsis form of the formula really only definitely indicates the start and end points and shows the general nature of interim points; it doesn't set a minimum number of terms or indeed the maximum number of terms - you are left to infer the exact quantity. In this case, there are $n$ terms in each sum, one each for all the numbers from $n+1$ to $2n$. $\endgroup$ – Joffan Feb 19 '15 at 0:30
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what would be p(n+1)? think about where that falls in your solution. p(n+1) is 1/(2n+2)

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Consider $P(n+1) - P(n)$. This difference has only three terms, all the middle terms cancel. And looking at those three terms you can easily show $P(n+1)-P(n)>0$. That proves that if $P(n)>7/12$ then $P(n+1)$ must also be.

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  • $\begingroup$ so is my answer wrong then? I have a midterm coming up soon and I'm confused on this subject. Would p(n+1) consist of 1/[(n+1)+1]+ 1/[(n+1)+2]+ 1/2(n+1) and p(n) be the regular 1/(n+1)+ 1/(n+2)+ 1/2n and you're Saying to subtract the two equations $\endgroup$ – user212401 Feb 19 '15 at 0:26
  • $\begingroup$ Yes your answer is not correct. Note how the question contains "$\dots$" but you seem to have lost them in your answer and they make all the difference. $\endgroup$ – Gregory Grant Feb 19 '15 at 0:35
  • $\begingroup$ Since P(2) is the statement $$\frac{1}{2 + 1} + \frac{1}{2 \cdot 2} = \frac{1}{3} + \frac{1}{4} = \frac{4}{12} + \frac{3}{12} = \frac{7}{12} \geq \frac{7}{12}$$ the assumption should be that $P(n) \geq \frac{7}{12}$. However, $P(n) > \frac{7}{12}$ for $n > 2$ for the reason you cited. $\endgroup$ – N. F. Taussig Feb 19 '15 at 0:38
  • $\begingroup$ So what would be p(n+1) in this case? $\endgroup$ – user212401 Feb 19 '15 at 3:09
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The three dots $\ldots$ are an ellipsis, indicating continuation of the preceding trend of terms up to the final term.

There are only two terms in the $n=2$ case. The ellipsis form of the formula really only definitely indicates the start point, the end point and the general nature of interim points; it doesn't set a minimum number of terms or indeed the maximum number of terms - you are left to infer the exact quantity.

In this case, there are $n$ terms in each sum, one each for all the numbers from $n+1$ to $2n$. So for the sake of clarity, here are a few terms: $$\begin{align} \frac{1}{3} + \frac{1}{4} &= \frac{7}{12} \tag{n=2} \\[0.5em] \frac{1}{4} + \frac{1}{5} + \frac{1}{6} &= \frac{37}{60} \tag{n=3} \\[0.5em] \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} &= \frac{533}{840} \tag{n=4} \\[0.5em] \end{align}$$

You can see that that $\frac{1}{3}$ has been replaced by $\frac{1}{5} + \frac{1}{6}$ in going from $n=2$ to $n=3$, and $\frac{1}{4}$ has been replaced by $\frac{1}{7} + \frac{1}{8}$ in going from $n=3$ to $n=4$. This should indicate the correct line of the induction argument.

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  • $\begingroup$ Okay so does p(n+1) in this question consist of 1/(n+1) + 1/(n+2)+ 1/2n? @joffan $\endgroup$ – user212401 Feb 19 '15 at 1:54
  • $\begingroup$ And p(n) is just the first term of 1/(n+1) +1/2n $\endgroup$ – user212401 Feb 19 '15 at 1:55
  • $\begingroup$ No, it's the sum of one over all the numbers between $n+1$ and $2n$ inclusive: $$p(n) = \sum_{k=n+1}^{2n} \frac{1}{k} $$ ... if we want to be formal about it. $\endgroup$ – Joffan Feb 19 '15 at 5:12

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