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Worked solution

Compare positive denominators to prove a strict lower bound

Given ∑r=1n1/(r+1+r)=n+1−1\sum_{r=1}^n1/(\sqrt{r+1}+\sqrt r)=\sqrt{n+1}-1, deduce ∑r=1n1/r>2(n+1−1)\sum_{r=1}^n1/\sqrt r>2(\sqrt{n+1}-1). marks

[2013/JJC/I/3(modified)]

CJC · Review problems · Question 9

Answer

∑r=1n1/r>2(n+1−1)\sum_{r=1}^n1/\sqrt r>2(\sqrt{n+1}-1).

Compare each positive reciprocal and sum

Since r<r+1\sqrt r<\sqrt{r+1}, 2r<r+1+r2\sqrt r<\sqrt{r+1}+\sqrt r. Taking positive reciprocals reverses the inequality: 1/(2r)>1/(r+1+r)1/(2\sqrt r)>1/(\sqrt{r+1}+\sqrt r). Summing from 11 to nn gives 12∑1/r>n+1−1\tfrac12\sum1/\sqrt r>\sqrt{n+1}-1; multiplying by 22 proves the result.

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Common mistakes

The inequality reverses when taking reciprocals of positive quantities.

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