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Prove by induction: $$\frac1{n+1} + \frac1{n+2} + \cdots +\frac1{3n+1} > 1$$

adding $1/(3m+4)$ as the next $m+1$ value proves pretty fruitless. Can I make some simplifications in the inequality that because the $m$ step is true by the inductive hypothesis, the 1 is already less than all those values?

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  • $\begingroup$ Please learn a bit of LaTeX and use it. And when you do write inline fractions like $1/(n+1)$, don't omit the parentheses. It will confuse your readers no end. I fixed it for you this time around. $\endgroup$ – Harald Hanche-Olsen Feb 25 '15 at 6:37
  • $\begingroup$ I agree with @Regret's comment--the general term is not at all obvious. My first inclination was to express the LHS using $\sum$-notation, but I can't do this without knowing how the terms look in general. $\endgroup$ – Daniel W. Farlow Feb 25 '15 at 6:39
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    $\begingroup$ You are adding three instead of one terms when you proceeding to $(m+1)$. $\endgroup$ – Vim Feb 25 '15 at 6:42
  • $\begingroup$ Possible duplicate of Prove that $1<\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1}$ $\endgroup$ – Martin Sleziak May 15 '18 at 8:47
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Hint: When you increase $n$ by $1$, the sum loses one term and gains three. What is the sum of the three gained terms minus the lost one?

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  • $\begingroup$ hmm. Are you referring to something like: 1/(3n+2)+1/(3n+3)+1/(3n+4)−1/(n+1) $\endgroup$ – rawrrawr Feb 25 '15 at 15:18
  • $\begingroup$ That's it, indeed. The sum of the middle positive term and the one negative term is easily simplified, so start there. You can get fancy and use the AM-HM inequality, or you can get the needed inequality by hand. $\endgroup$ – Harald Hanche-Olsen Feb 25 '15 at 18:07
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More generally (one of my favorite phrases), let $s_k(n) =\sum\limits_{i=n+1}^{kn+1} \frac1{i} $.

I will show that $s_k(n+1)>s_k(n)$ for $k \ge 3$.

In particular, for $n \ge 1$ $s_3(n) \ge s_3(1) =\frac1{2}+\frac1{3}+\frac1{4} =\frac{6+4+3}{12} =\frac{13}{12} > 1 $.

$\begin{array}\\ s_k(n+1)-s_k(n) &=\sum\limits_{i=n+2}^{kn+k+1} \frac1{i}-\sum\limits_{i=n+1}^{kn+1} \frac1{i}\\ &=\sum\limits_{i=n+2}^{kn+1} \frac1{i}+\sum\limits_{i=kn+2}^{kn+k+1} \frac1{i} -\left(\frac1{n+1}+\sum\limits_{i=n+2}^{kn+1} \frac1{i}\right)\\ &=\sum\limits_{i=kn+2}^{kn+k+1} \frac1{i}-\frac1{n+1}\\ &=\sum\limits_{i=2}^{k+1} \frac1{kn+i}-\frac1{n+1}\\ &=\frac1{kn+2}+\frac1{kn+k+1}+\sum\limits_{i=3}^{k} \frac1{kn+i}-\frac1{n+1}\\ \end{array} $

$\sum\limits_{i=3}^{k} \frac1{kn+i} \ge \sum\limits_{i=3}^{k} \frac1{kn+k} = \frac{k-2}{kn+k} $.

If we can show that $\frac1{kn+2}+\frac1{kn+k+1} \ge \frac{2}{kn+k} $, then $s_k(n+1)-s_k(n) \ge \frac{2}{kn+k}+\frac{k-2}{kn+k}-\frac1{n+1} = \frac{k}{kn+k}-\frac1{n+1} = \frac{1}{n+1}-\frac1{n+1} =0 $.

But

$\begin{array}\\ \frac1{kn+2}+\frac1{kn+k+1}-\frac{2}{kn+k} &=\frac{kn+k+1+(kn+2)}{(kn+2)(kn+k+1)}-\frac{2}{kn+k}\\ &=\frac{2kn+k+3}{(kn+2)(kn+k+1)}-\frac{2}{kn+k}\\ &=\frac{(2kn+k+3)(kn+k)-2(kn+2)(kn+k+1)}{(kn+2)(kn+k+1)(kn+k)}\\ \end{array} $

Looking at the numerator,

$\begin{array}\\ (2kn+k+3)(kn+k)-2(kn+2)(kn+k+1) &=2k^2n^2+kn(k+3+2k)+k(k+3)\\ &-2(k^2n^2+kn(k+3)+2(k+1)\\ &=2k^2n^2+kn(3k+3)+k(k+3)\\ &-2k^2n^2-2kn(k+3)-4(k+1)\\ &=kn(3k+3)+k(k+3)\\ &-kn(2k+6)-4(k+1)\\ &=kn(k-3)+k(k+3)-4(k+1)\\ &=kn(k-3)+k^2-k-4\\ &> 0 \quad\text{for $k \ge 3$} \end{array} $

and we are done.

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Hint: use that $$\frac1{3n+2}+\frac1{3n+3}+\frac1{3n+4}-\frac1{n+1}=\cdots >0.$$

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