Let's prove this inequality, with/without induction $\sum\limits_{i=1}^n\frac1i \le \sqrt n$

Question

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.

Answer 1

hint: $\dfrac{1}{i} \le \dfrac{1}{\sqrt{i}+\sqrt{i-1}} = \sqrt{i} - \sqrt{i-1}, i \ge 4$.

Answer 2

Another hint: harmonic series is bounded by $\ln{n}+1$, which is $\ln{n}+1 < \sqrt{n}$ at least because $$\lim_{n \rightarrow \infty} \frac{1+\log{n}}{\sqrt{n}}=0$$

Answer 3

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 $n\ge7$

$$\begin{align} 1+{1\over2}+{1\over3}+\cdots+{1\over n} &\le1+{1\over2}+{1\over3}+{1\over4}+{1\over5}+{1\over6}+{1\over7}+\int_7^n{dx\over x}\\ &={363\over140}+\int_7^n{dx\over x}\\ &\le{363\over140}+\int_7^n{dx\over2\sqrt x }\\ &={363\over140}+\sqrt n-\sqrt7\\ &\le\sqrt n\qquad\text{since }\left(363\over140\right)^2\approx6.7229\lt7 \end{align}$$

Answer 4

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.

Answer 5

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.

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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?

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