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Doing proof by induction exercises with inequalities, I got stuck on one that is a bit different from the others. There is no $n$ term on the rightmost part of the inequality:

Prove that the following holds for all $n \ge 1$:

$$\frac{1}{n + 1}+\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}\le \frac{5}{6}$$

All the other proofs I did before had some $n$ involved there, but now there is none. I wonder how will this difference affect my standard attempt:


Test for $n = 1$:

$$\frac{1}{3} \le \frac{5}{6} \implies 6 \le 15$$


We have to prove that it holds for $n + 1$, that is:

$$\frac{1}{n + 1}+\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}+\frac{1}{(n + 1)+(n+2)}\le \frac{5}{6}$$


Asssume $$\frac{1}{n + 1}+\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}\le \frac{5}{6}$$


We start off with

$$\frac{1}{n + 1}+\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}+\frac{1}{(n + 1)+(n+2)}\le \frac{5}{6}$$

And now I see the problem. Normally, I apply the hypothesis first on this part:

$$\color{blue}{\frac{1}{n + 1}+\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}}+\frac{1}{(n + 1)+(n+2)}\le \frac{5}{6}$$

But all it yields is this:

$$\color{blue}{\left ( \textrm{something smaller or equal than } \frac{5}{6} \right )} + \frac{1}{(n + 1)+(n+2)}\le \frac{5}{6}$$

If I move $\frac{1}{(n + 1)+(n+2)}$ to the other side, all I get is

$$\color{blue}{\left ( \textrm{something smaller or equal than } \frac{5}{6} \right )} \le \frac{5}{6} - \frac{1}{(n + 1)+(n+2)}$$

Which cannot be guaranteed.

How can I prove this, then?

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  • $\begingroup$ If it is not compulsory to use induction... are you familiar with A.M-H.M inequality? $\endgroup$ – user87543 Dec 3 '13 at 4:26
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    $\begingroup$ The first proof that comes to my mind doesn’t use induction. $$\sum_{k=n+1}^{2n+1}\frac1k<\int_n^{2n+1}\frac{dx}x=\ln(2n+1)-\ln n=\ln\left(2+\frac1n\right)\;,$$ which is less than $\frac56$ for all $n\ge 4$. Then just check the result for $n=1,2,3$. $\endgroup$ – Brian M. Scott Dec 3 '13 at 4:27
  • $\begingroup$ @PraphullaKoushik: Well, it's an induction exercise, so I should use it. I don't recognise that term for inequality (or perhaps I do, but in Spanish) - I will take a look at it, thanks. $\endgroup$ – Zol Tun Kul Dec 3 '13 at 4:30
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    $\begingroup$ The statement that you're trying to prove in the induction step is not quite correct. The sum ought to begin with the term $\frac{1}{n+2}$. $\endgroup$ – Sammy Black Dec 3 '13 at 4:32
  • $\begingroup$ Ok.. I am sure it would be easy with that inequality but i will try with induction also $\endgroup$ – user87543 Dec 3 '13 at 4:33
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Let $$a_n=\sum_{k=n+1}^{2n+1}\frac1k\;,$$ and show that the sequence $\langle a_n:n\ge 1\rangle$ is decreasing. There’s more in the spoiler-protected block below.

$\displaystyle\sum_{k=n+1}^{2n+1}\frac1k-\sum_{k=n+2}^{2n+3}\frac1k=\frac1{n+1}-\frac1{2n+2}-\frac1{2n+3}=\frac1{2n+2}-\frac1{2n+3}=\ldots$

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we assume

$$\frac{1}{n + 1}+\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}\le \frac{5}{6}$$

looking for $n+1$ case, we see : $$\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}+\frac{1}{(n+1)+(n+2)}=\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}+\frac{1}{n+1}-\frac{1}{n+2}=\frac{1}{n + 1}+\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}-\frac{1}{n+2}\leq \frac{5}{6}-\frac{1}{n+2}\leq \frac{5}{6}$$

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