Let's prove this inequality, with/without induction $\sum\limits_{i=1}^n\frac1i \le \sqrt n$ Let's prove this inequality, using with induction (or without), for$\quad n\ge 7$ , $\quad \boxed{\displaystyle\sum\limits_{i=1}^n\dfrac1i \le \sqrt n}$
My Attempt:
Since ;$\quad\displaystyle\sum\limits_{i=1}^{n+1}\dfrac1i=\displaystyle\sum_{i=1}^n\dfrac1i+\dfrac1{1+n}\le \sqrt n +\dfrac1{1+n}$
We need a prove this,
$$\sqrt n +\dfrac1{1+n}\le \sqrt{n+1}$$
Therefore,
$$\dfrac1{1+n}\le \sqrt{n+1}-\sqrt n$$
Hence,
$$\dfrac1{\sqrt{n+1}-\sqrt n}\le 1+n$$
So,
$$\sqrt{n+1}+\sqrt n\le 1+n$$
But, I couldn't complete this.
 A: hint:  $\dfrac{1}{i} \le \dfrac{1}{\sqrt{i}+\sqrt{i-1}} = \sqrt{i} - \sqrt{i-1}, i \ge 4$. 
A: Another hint: harmonic series is bounded by $\ln{n}+1$, which is $\ln{n}+1 < \sqrt{n}$ at least because $$\lim_{n \rightarrow \infty} \frac{1+\log{n}}{\sqrt{n}}=0$$ 
A: Note that ${1\over x}\le{1\over2\sqrt x}$ for $x\ge4$.  Combined with the standard integral comparison for the harmonic series, this lets us conclude that, for $n\ge7$
$$\begin{align}
1+{1\over2}+{1\over3}+\cdots+{1\over n}
&\le1+{1\over2}+{1\over3}+{1\over4}+{1\over5}+{1\over6}+{1\over7}+\int_7^n{dx\over x}\\
&={363\over140}+\int_7^n{dx\over x}\\
&\le{363\over140}+\int_7^n{dx\over2\sqrt x }\\
&={363\over140}+\sqrt n-\sqrt7\\
&\le\sqrt n\qquad\text{since }\left(363\over140\right)^2\approx6.7229\lt7
\end{align}$$
A: For each $1\leq k\leq n$,
we have
$\frac{1}{k}\leq \int_{k-1}^k \frac{dt}{t}$
by addition, we get
$\sum_{k=2}^n \frac{1}{k}\leq \int_1^n\frac{dt}{t}=\log(n)$
and
$\sum_{k=1}^n\frac{1}{k}\leq 1+\log(n)$.
but
$x\mapsto 1+\log(x)-\sqrt{x}$ is decreasing at $[7,+\infty)$ by derivative.
so
$1+\log(x)\leq \sqrt{x}$. 
qed.
A: Let us say we want to use the known inequality1 $$\sum_{k=1}^n \frac1{2\sqrt k} \le \sqrt n.$$
Since for $k\ge 4$ we have $k\ge 2\sqrt k$ and $$\frac 1k \le \frac1{2\sqrt k},$$ it suffices to find $n_0\ge 4$ such that $$\sum_{k=1}^{n_0} \frac 1k \le \sum_{k=1}^{n_0} \frac1{2\sqrt k}$$ and then we get
$$ \sum_{k=1}^n \frac1k = \sum_{k=1}^{n_0} \frac1k + \sum_{k=n_0+1}^n \frac1k \le  \sum_{k=1}^{n_0} \frac1{2\sqrt k} + \sum_{k=n_0+1}^n \frac1{2\sqrt k} = 
\sum_{k=1}^n \frac1{2\sqrt k} \le \sqrt n.$$
Still the part which remains to be done manually is to find such $n_0$. According to WolframAlpha $n_0=10$ suffices.

1 There are several posts on this site with proofs of this inequality, for example:


*

*Use induction to prove that $ 1 + \frac {1}{\sqrt{2}} + \frac {1}{\sqrt{3}} .... + \frac {1}{\sqrt{n}} < 2\sqrt{n}$

*Induction: show that $\sum\limits_{k=1}^n \frac{1}{\sqrt{k}} < 2 \sqrt{n}$ for all n $\in Z_+$

*Prove that $\sqrt{n} \le \sum_{k=1}^n \frac{1}{\sqrt{k}} \le 2 \sqrt{n} - 1$ is true for $n \in \mathbb{N}^{\ge 1}$

*Can the inequality $\sum\limits_{i=1}^n\frac{1}{\sqrt{i}} < 2\sqrt{n} - 1$ be proved without induction?

This is along the same lines as the hint posted in this answer, just with more details.
