# Inequality $\frac{1}{n+1}+\frac{1}{n+2}+…+\frac{1}{3n+1} \ge 1$

I have this inequality that I don't know how to prove. Possibly the inequality between means might be useful. For $n \in \mathbb{N}$:

$$\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1} \ge 1$$

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if you want to use inequalities between means: use the inequality between arithmetic and harmonic mean and compute/estimate the harmonic mean of $1/(n+1),\dots,1/(3n+1)$ – user8268 Oct 28 '12 at 19:25
Try using mathematical induction. – Jonah Sinick Oct 28 '12 at 20:10

It follows immediately from arithmetic mean harmonic mean inequality. Recall that given a set of positive numbers, $\{a_k\}_{k=1}^{k=m}$, we have that $$\dfrac{\displaystyle\sum_{k=1}^{m} a_k}m \geq \dfrac{m}{\displaystyle \sum_{k=1}^m \dfrac1{a_k}}$$ $$\dfrac{\displaystyle \sum_{k=1}^{2n+1} \dfrac1{n+k}}{2n+1} \geq \dfrac{2n+1}{\displaystyle \sum_{k=1}^{2n+1} (n+k)}$$ $$\displaystyle \sum_{k=1}^{2n+1} (n+k) = n(2n+1) + \dfrac{(2n+1)(2n+2)}2 = (2n+1)^2$$ Hence, $$\dfrac{\displaystyle \sum_{k=1}^{2n+1} \dfrac1{n+k}}{2n+1} \geq \dfrac{2n+1}{(2n+1)^2}$$ Hence, $$\displaystyle \sum_{k=1}^{2n+1} \dfrac1{n+k} \geq 1$$

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Write this as $$\sum_{j=1}^{2n+1} \dfrac{1}{n+j} \ge \int_1^{2n+2} \dfrac{dx}{n+x} = \ln \left(\frac{3n+2}{n+1}\right)$$

Now $\dfrac{3n+2}{n+1} \ge e$ for $n > \dfrac{e-2}{3-e} \approx 2.549$. Do the cases $n=1$ and $2$ separately.

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We induct on $n$. For $n=1$, it's trivial. Now assume it holds for $n=k$, then $$\frac{1}{(k+1)+1}+\cdots+\frac{1}{3(k+1)}+\frac{1}{3(k+1)+1}=\frac{1}{k+2}+\cdots+\frac{1}{3k+4}$$ $$=(\frac{1}{k+1}+\cdots+\frac{1}{3k+1})-\frac{1}{k+1}+\frac{1}{3k+2}+\frac{1}{3k+3}+\frac{1}{3k+4}$$ Since the first expression is at least $1$ by the inductive hypothesis, it suffices to show $$\frac{1}{3k+2}+\frac{1}{3k+3}+\frac{1}{3k+4}\ge \frac{1}{k+1}$$ But by Cauchy-Schwarz we obtain $$(3k+2+3k+3+3k+4)(\frac{1}{3k+2}+\frac{1}{3k+3}+\frac{1}{3k+4})\ge9$$ Which is equivalent to what we want after dividing by $9(k+1)$. So the inductive hypothesis always holds.

We can also use user8268's hint and apply Cauchy or AM-HM directly: $$(k+1+\cdots+3k+1)(\frac{1}{k+1}+\cdots+\frac{1}{3k+1})\ge (2k+1)^2$$ So it suffices to show $$(2k+1)^2\ge (k+1+\cdots+3k+1)=(1+\cdots+3k+1)-(1+\cdots+k)$$ or $$4k^2+4k+1\ge \frac{(3k+1)(3k+2)}{2}-\frac{k(k+1)}{2}=\frac{8k^2+8k+2}{2}=4k^2+4k+1$$ but this is clearly true.

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I tried this method as well but the second inequality is not true. Let's put k=6, then we get: $$L=\frac{1}{20}+\frac{1}{21}+\frac{1}{22} \approx 0.14$$ while the right side is $$R=\frac{1}{7} \approx 0.16$$ – Max Oct 28 '12 at 19:43
Actually, $L=\frac{1}{20}+\frac{1}{21}+\frac{1}{22}\approx 0.1431$, while $R=\frac{1}{7}\approx 0.1429$, so I believe the inequality is true. – pi37 Oct 28 '12 at 19:54
Yes indeed, you are right. Actually i did this exercise the same way you have, but instead of $\frac{1}{k+1}$ I, somehow, got $\frac{1}{k}$ for which this inequality is not true. Thanks! – Max Oct 28 '12 at 20:08

In general, if $f$ is a convex function and $p_j$ are non-negative coefficients such that $\sum p_j = 1$ then

$$\sum_j p_j \, f(x_j) \geq f(\sum_j p_j \, x_j).$$

(This is Jensen's inequality.) Take $f(x) = 1/x$, $p_j = 1/(2n + 1)$ and $x_j = n + j$ for $j \in \{1, \dotsc, 2n+1 \}$ and your inequality follows.

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