How do I prove $\frac 34\geq \frac{1}{n+1}+\frac {1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{n+n}$

Solution 1:

You don't need calculus here (although it would make things easier). Let $$S=\frac{1}{n+1}+\frac {1}{2n}+\frac{1}{n+3}+...+\frac{1}{2n}$$

and write it both ways, à la Gauss:

$$\begin{align} 2S &= \frac{1}{n+1}+\frac {1}{n+2}\,\,\,+\,\,\frac{1}{n+3}\,+...+\,\,\,\frac{1}{2n} \\ &\,\,\,\,+\frac{1}{2n}\,\,\;+\frac {1}{2n-1}+\frac{1}{2n-2}+...+\frac{1}{n+1}\\ &=\sum_{k=1}^n\left(\frac{1}{n+k}+\frac{1}{2n-(k-1)}\right)\\ &<\sum_{k=1}^n\left(\frac{1}{n+k}+\frac{1}{2n-k}\right)\\ &=\sum_{k=1}^n\frac{3n}{(n+k)(2n-k)}\\ &\le\sum_{k=1}^n\frac{3n}{2n^2} \,\,\,\,\text{because}\, (n+k)(2n-k) \ge 2n^2 \text{ if } 1 \le k \le n\\ &=\frac{3}{2n}\sum_{k=1}^n1\\ &=\frac32 \end{align}$$

Hence $S < \dfrac{3}{4}$.

Solution 2:

A variation on the inequality that @TonyK already mentioned. For $1\leq k\leq n$ $$2n^2\leq 2n^2+k(n-k) = (n+k)(2n-k)$$ and so, dividing both sides by $2n^2(n+k)$, $$\frac1{n+k}\leq\frac1n-\frac{k}{2n^2}.$$ Then $$\sum_{k=1}^n\frac1{n+k}\leq 1 - \sum_{k=1}^n\frac{k}{2n^2}=1-\frac{n(n+1)}{4n^2}=\frac34-\frac1{4n}.$$

Solution 3:

Your sum can be expressed as a Riemann Sum. Specifically

$$\frac 1{n+1}+\dots+\frac 1{2n}<\int_n^{2n}\frac {1}{x}dx=ln(2n)-ln(n)=ln(2)\sim.693$$

Solution 4:

$f(x)=\frac{1}{n+x}$ is a convex function on $\mathbb{R}^+$, hence:

$$ \frac{f(1)+f(2)+\ldots+f(n)}{n}\geq f\left(\frac{1+2+\ldots+n}{n}\right) \tag{1} $$ implies:

$$ \frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{n+n}\geq \frac{n}{n+\frac{1+2+\ldots+n}{n}}=\frac{2n}{3n-1}.\tag{2}$$ On the other hand, the Hermite-Hadamard inequality gives: $$ \frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{n+n}\leq \int_{n+\frac{1}{2}}^{2n+\frac{1}{2}}\frac{dx}{x} = \log\left(\frac{4n+1}{2n+1}\right)\leq \log 2.\tag{3}$$

An alternative approach is to prove first that the sequence $\{a_n\}_{n\geq 1}$ given by $a_n=\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2n}$ is increasing, then to compute its limit at $n\to +\infty$ and check it is less than $\frac{3}{4}$.