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How do I prove the following inequality

$$\frac 34\geq \frac{1}{n+1}+\frac {1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{n+n}$$

without the help of induction?

Thanks for any help!!

user26857
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Soham
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5 Answers5

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You don't need calculus here (although it would make things easier). Let $$S=\frac{1}{n+1}+\frac {1}{2n}+\frac{1}{n+3}+...+\frac{1}{2n}$$

and write it both ways, à la Gauss:

$$\begin{align} 2S &= \frac{1}{n+1}+\frac {1}{n+2}\,\,\,+\,\,\frac{1}{n+3}\,+...+\,\,\,\frac{1}{2n} \\ &\,\,\,\,+\frac{1}{2n}\,\,\;+\frac {1}{2n-1}+\frac{1}{2n-2}+...+\frac{1}{n+1}\\ &=\sum_{k=1}^n\left(\frac{1}{n+k}+\frac{1}{2n-(k-1)}\right)\\ &<\sum_{k=1}^n\left(\frac{1}{n+k}+\frac{1}{2n-k}\right)\\ &=\sum_{k=1}^n\frac{3n}{(n+k)(2n-k)}\\ &\le\sum_{k=1}^n\frac{3n}{2n^2} \,\,\,\,\text{because}\, (n+k)(2n-k) \ge 2n^2 \text{ if } 1 \le k \le n\\ &=\frac{3}{2n}\sum_{k=1}^n1\\ &=\frac32 \end{align}$$

Hence $S < \dfrac{3}{4}$.

Glorfindel
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TonyK
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A variation on the inequality that @TonyK already mentioned. For $1\leq k\leq n$ $$2n^2\leq 2n^2+k(n-k) = (n+k)(2n-k)$$ and so, dividing both sides by $2n^2(n+k)$, $$\frac1{n+k}\leq\frac1n-\frac{k}{2n^2}.$$ Then $$\sum_{k=1}^n\frac1{n+k}\leq 1 - \sum_{k=1}^n\frac{k}{2n^2}=1-\frac{n(n+1)}{4n^2}=\frac34-\frac1{4n}.$$

WimC
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Your sum can be expressed as a Riemann Sum. Specifically

$$\frac 1{n+1}+\cdots+\frac 1{2n}<\int_n^{2n}\frac {1}{x}~\mathrm dx=\ln(2n)-\ln(n)=\ln(2)\sim.693.$$

User1865345
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lulu
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  • I am too confident with calculus...it would be helpful if the answer is a little more explanatory....thanks.. – Soham Feb 06 '16 at 09:50
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    The definite integral is the area under the curve for $n≤x≤2n$. So...draw a picture. the graph of $f(x)=\frac 1x$ is a strictly decreasing function of $x$. divide the interval $n≤x≤2n$ by putting points at ${n+1,n+2,\dots,n+n}$ and approximate the area by drawing rectangles with height given by the right hand endpoints. so you have rectangles with area ${\frac 1{n+1}, \frac 1{n+2},\dots,\frac 1{n+n}}$. If you've drawn the picture you'll see that each of these rectangles, hence the sum of their areas, lies entirely under the graph, so their combined area is less than the total area. – lulu Feb 06 '16 at 09:55
  • @lulu What do you mean by total area and combined area...please? – Soham Feb 06 '16 at 10:05
  • Do you know any integral calculus? This is a very basic example. See, for example the usual (calculus based) proof of the divergence of the sum $\frac 11 +\frac 12+\dots$. You can find that proof in any first year calculus text. Or you can read about it here https://en.wikipedia.org/wiki/Harmonic_series_(mathematics) I note that the wikipefia article includes a picture of the region. But, if you've never seen any calculus, this might not be the best place to start. – lulu Feb 06 '16 at 10:08
  • I am a beginner in calculus...not too proficient in it.....I know the basics of calculus....I could not understand what you were saying as you had not included a figure....now I can understand it...thanks for your help... – Soham Feb 07 '16 at 01:43
  • Had an exciting time learning Riemann sum and integrals....I accepted your answer after so many days because I had to utilise the time in learning Riemann sums....thanks a lot!! – Soham Feb 12 '16 at 09:42
  • No problem. Fascinating stuff, no? – lulu Feb 12 '16 at 11:56
  • Yeah...really... – Soham Feb 13 '16 at 03:07
4

Expanding on the idea of the other answer:

$$\frac1{n+1}+\frac1{n+2}+\ldots+\frac1{2n}=\frac1n\left(\frac1{1+\frac1n}+\frac1{1+\frac2n}+\ldots+\frac1{1+\frac nn}\right)\rightarrow$$

$$\rightarrow\int_0^1\frac{dx}{1+x}=\log2<\frac34$$

DonAntonio
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$f(x)=\frac{1}{n+x}$ is a convex function on $\mathbb{R}^+$, hence:

$$ \frac{f(1)+f(2)+\ldots+f(n)}{n}\geq f\left(\frac{1+2+\ldots+n}{n}\right) \tag{1} $$ implies:

$$ \frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{n+n}\geq \frac{n}{n+\frac{1+2+\ldots+n}{n}}=\frac{2n}{3n-1}.\tag{2}$$ On the other hand, the Hermite-Hadamard inequality gives: $$ \frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{n+n}\leq \int_{n+\frac{1}{2}}^{2n+\frac{1}{2}}\frac{dx}{x} = \log\left(\frac{4n+1}{2n+1}\right)\leq \log 2.\tag{3}$$

An alternative approach is to prove first that the sequence $\{a_n\}_{n\geq 1}$ given by $a_n=\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2n}$ is increasing, then to compute its limit at $n\to +\infty$ and check it is less than $\frac{3}{4}$.

Jack D'Aurizio
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