How do I prove $\frac 34\geq \frac{1}{n+1}+\frac {1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{n+n}$ How do I prove the following inequality
$$\frac 34\geq \frac{1}{n+1}+\frac {1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{n+n}$$
without the help of induction?
Thanks for any help!!
 A: 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 $S < \dfrac{3}{4}$.
A: 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 $$\sum_{k=1}^n\frac1{n+k}\leq 1 - \sum_{k=1}^n\frac{k}{2n^2}=1-\frac{n(n+1)}{4n^2}=\frac34-\frac1{4n}.$$
A: Your sum can be expressed as a Riemann Sum.  Specifically 

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

A: $f(x)=\frac{1}{n+x}$ 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 $\frac{3}{4}$.
A: Expanding on the idea of the other answer:
$$\frac1{n+1}+\frac1{n+2}+\ldots+\frac1{2n}=\frac1n\left(\frac1{1+\frac1n}+\frac1{1+\frac2n}+\ldots+\frac1{1+\frac nn}\right)\rightarrow$$
$$\rightarrow\int_0^1\frac{dx}{1+x}=\log2<\frac34$$
