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I have proven the left hand side inequality. Can someone give me a hint for the right hand side ?

Proof for the left hand side: $\sum_{r=1}^{n} \frac{1}{n+r} >\sum_{r=1}^{n} \frac{1}{2n}$.

Aditya
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    Here is a possible proof which is clearly not the one the author of the problem had in mind: $$ \sum\limits_{r = 1}^n {\frac{1}{{n + r}}} \le \sum\limits_{r = 1}^n {\int_{n + r - 1}^{n + r} {\frac{{dx}}{x}} } = \int_n^{2n} {\frac{{dx}}{x}} = \log 2 = 0.693 \ldots < \frac{3}{4}. $$ – Gary Jun 15 '20 at 20:04
  • Also: https://math.stackexchange.com/q/1974177/42969, https://math.stackexchange.com/q/1974167/42969, https://math.stackexchange.com/q/2418810/42969 – Martin R Jun 15 '20 at 20:17

2 Answers2

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As commented but perhaps slightly more directly:

$$\sum_{r=1}^n\frac1{n+r}=\frac1n\sum_{r=1}^n\frac1{1+\frac rn}\le\lim_{n\to\infty}\frac1n\sum_{r=1}^n\frac1{1+\frac rn}=\int_0^1\frac{dx}{1+x}=\ln2\le\frac34$$

The inquality above is due to the fact that all the terms in the sums are non-negative and in fact its sequence of partial sums is a monotone increasing sequence.

DonAntonio
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  • why bother with integral at all if the sum is the difference between $H_{2n}$ and $H_n$ and, since you take the limit anyway, it's $\log 2$? – Alex Jun 15 '20 at 20:27
  • Because I think it's a very nice example of using Riemann Sums to evaluate, or in this case to bound, in the limits. – DonAntonio Jun 15 '20 at 22:46
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By C-S $$\sum_{r=1}^n\frac{1}{n+r}=1+\sum_{r=1}^n\left(\frac{1}{n+r}-\frac{1}{n}\right)=1-\frac{1}{n}\sum_{r=1}^n\frac{r}{n+r}=$$ $$=1-\frac{1}{n}\sum_{r=1}^n\frac{r^2}{nr+r^2}\leq1-\frac{\left(\sum\limits_{r=1}^nr\right)^2}{n\sum\limits_{r=1}^n(nr+r^2}=1-\frac{\frac{n^2(n+1)^2}{4}}{n\left(\frac{n^2(n+1)}{2}+\frac{n(n+1)(2n+1)}{6}\right)}=$$ $$=\frac{7n-1}{2(5n+1)}\leq\frac{3}{4}.$$

Michael Rozenberg
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