Mathematical induction for inequalities: $\frac1{n+1} + \frac1{n+2} + \cdots +\frac1{3n+1} > 1$

Question

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?

Answer 1

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?

Answer 2

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.

Answer 3

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