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Show that $$1+\dfrac12+\dfrac13+\text{...}+\dfrac{1}{2n-2}+\dfrac{1}{2n-1}<n$$ for all $n>1,n\in N.$

Initially, we should prove the proposition is true for $n=2$. I wasn't sure how the LHS was supposed to look, as the last term when $n=2$ is equal to $$\dfrac{1}{2n-1}=\dfrac{1}{2\cdot2-1}=\dfrac{1}{3}$$ Should we also include $1$ and $\dfrac12$? Well, I continued the solution with including them, so $$1+\dfrac12+\dfrac13\overset{?}{<}2\\\dfrac{11}{6}\overset{?}{<}2$$ which is obviously true. Now let's suppose that the inequality holds for $n=k\ge2,$ $$1+\dfrac12+\dfrac13+\text{...}+\dfrac{1}{2k-2}+\dfrac{1}{2k-1}<n$$ Then it must also be true for $n=k+1$, so we have to prove $$1+\dfrac12+\dfrac13+\text{...}+\dfrac{1}{2k}+\dfrac{1}{2k+1}\overset{?}{<}k+1\\1+\dfrac12+\dfrac13+\text{...}+\dfrac{1}{2k-2}+\dfrac{1}{2k-1}+\dfrac{1}{2k}+\dfrac{1}{2k+1}\overset{?}{<}k+1$$ How do we do that?

kormoran
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    What is the difference of the sum when $n=k$ and when $n=k+1?$ – Thomas Andrews Jan 16 '22 at 22:01
  • "Initially, we should prove the proposition is true for $n=2$. I wasn't sure how the LHS was supposed to look, as the last term when $n=2$ is equal to $$\dfrac{1}{2n-1}=\dfrac{1}{2\cdot2-1}=\dfrac{1}{3}$$ Should we also include $1$ and $\dfrac12$? " Yes. But I'm interested in why you doubt this? $$$$ Maybe I can clarify why this is the case. When $n=2,\quad 1+\dfrac12+\dfrac13+\text{...}+\dfrac{1}{2n-2}+\dfrac{1}{2n-1} = 1+\dfrac12+\dfrac13 $ (by definition). – Adam Rubinson Jan 16 '22 at 22:06
  • $\dfrac{1}{2k}+\dfrac{1}{2k+1} < \dfrac12+\dfrac12=1$ when $k\ge 1$ – Henry Jan 17 '22 at 17:51

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You can get the inequality by estimating the sum of pairs of consecutive fractions, starting at the second one: $$ 1+\underbrace{\left(\frac12+\frac13\right)}_{< 1}+\cdots+\underbrace{\left(\frac{1}{2n-2}+\frac{1}{2n-1}\right)}_{< 1}< 1 + \underbrace{1 + \cdots + 1}_{n-1 \text{ terms}} = n $$

Regarding your proof by induction, in particular the induction step: If you assume that the inequality holds for $n=k \ge 2$ then you assume that $$ 1+\frac12+\frac13+\cdots+\frac{1}{2k-2}+\frac{1}{2k-1}<k \, . $$ It follows that $$ 1+\frac12+\frac13+\cdots+\frac{1}{2k-2}+\frac{1}{2k-1} +\frac{1}{2k}+\frac{1}{2k+1} \\ = \underbrace{\left(1+\frac12+\frac13+\cdots+\frac{1}{2k-2}+\frac{1}{2k-1}\right)}_{< k} +\underbrace{\left(\frac{1}{2k}+\frac{1}{2k+1}\right)}_{< 1} \\ < k + 1 = k+1 \, , $$ which means that the inequality holds for $n=k+1$.

Martin R
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