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

summation induction inequality harmonic-numbers

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?


Solution 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?

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