By mathematical induction prove that?

Question

$\frac{1}{n + 1} + \frac{1}{n + 2} + \frac{1}{n + 3} + ..... +\frac{1}{3(n) + 1} > 1 $;

Here in this sequence after checking the basis for n = 1 , i.e $\frac{1}{4} > 1$, Which cannot be true;

I think I am wrong at some point , If yes then how can we prove this sequence by mathematical induction?

Answer 1

For $n=1$ we need to check that $$\frac{1}{2}+\frac{1}{3}+\frac{1}{4}>1$$ or $$\frac{13}{12}>1.$$ Let $$\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1}>1.$$ Thus, $$\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{3n+1}+\frac{1}{3n+2}+\frac{1}{3n+3}+\frac{1}{3n+4}=$$ $$=\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1}+\frac{1}{3n+2}+\frac{1}{3n+3}+\frac{1}{3n+4}-\frac{1}{n+1}>$$ $$>1+\frac{1}{3n+2}+\frac{1}{3n+3}+\frac{1}{3n+4}-\frac{1}{n+1}=1+\frac{1}{3n+2}+\frac{1}{3n+4}-\frac{2}{3(n+1)}=$$ $$=1+\frac{2}{3(n+1)(3n+2)(3n+4)}>1$$ and we are done!