prove that $2(\sqrt{n+1}-1)<1+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{n}}< 2\sqrt{n}$

Question

prove $2(\sqrt{n+1}-1)<1+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{n}}< 2\sqrt{n}$ by mathematical indcution.

my attempt: we prove for $n=1$

for $n=1$ than $0.828<1<2$ so true for $n=1$

we assume that this is true for $n=k$

ie $2\sqrt{k+1}-1<1+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{k}}< 2\sqrt{k}$

Now we have to prove for $n=k+1$

consider $\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}< 2\sqrt{k}+\frac{1}{\sqrt{k+1}}\leq 2\sqrt{k}+1<2\sqrt{k+1}$ Is this correct?

what about otherside

@JitendraSingh..im asking how can we prove by using induction — Inverse Problem, Jul 15 '21 at 13:32
Not correct since squaring both sides gives the following: $$2\sqrt k + 1 < 2 \sqrt{k+1} \ 4k+4\sqrt{k} + 1 < 4k+4 \ \sqrt{k} < \frac{3}{4}$$ — VIVID, Jul 15 '21 at 13:36
When you say is this correct... I prefer to revert the question to you: have you proven that $2 \sqrt{k} + 1 \lt 2 \sqrt{k+1}?$ What makes you think that the inequality is valid? — mathcounterexamples.net, Jul 15 '21 at 13:37

score 2 · Answer 1 · answered Jul 15 '21 at 13:50

(Using mathematical induction),

(As you did) Now we assume that this is true for $n=k$, i.e.

$$2\sqrt{k+1}-1<1+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{k}}< 2\sqrt{k}\tag 1$$

Now we have to prove for $n=k+1$. All you have to do is to prove

$$2\sqrt{k+2}-1<2\sqrt{k+1}-1+\frac{1}{\sqrt{k+1}}\tag 2$$ and $$2\sqrt k+\frac{1}{\sqrt{k+1}}<2\sqrt{k+1}\tag 3$$

To prove $(2)$, $$2\sqrt{k+2}-2\sqrt{k+1}=\frac{2}{\sqrt{k+2}+\sqrt{k+1}}<\frac{2}{\sqrt{k+1}+\sqrt{k+1}}=\frac{1}{\sqrt{k+1}}$$ Therefore $(2)$ is true.

To prove $(3)$, $$2\sqrt{k+1}-2\sqrt k=\frac2{\sqrt{k+1}+\sqrt k}>\frac2{\sqrt{k+1}+\sqrt {k+1}}=\frac1{\sqrt {k+1}}$$ Therefore $(3)$ is true.

Now from $(1),(2)$ and $(3)$, the equation for $n=k+1$ also holds.

score 1 · Answer 2 · answered Jul 15 '21 at 13:55

Use the condiction from $n=k$ and what we should get is that $$2\sqrt{k+2}-1<\underbrace{\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{k}}}_{>2\sqrt{k+1}-1}+\frac{1}{\sqrt{k+1}}$$ So we just have to compare $2\sqrt{k+2}-1$ with $2\sqrt{k+1}+\frac{1}{\sqrt{k+1}}-1 $ $$(2\sqrt{k+2} )^2=4(k+2)$$ $$(2\sqrt{k+1}+\frac{1}{\sqrt{k+1}})^2=4(k+1)+4+\frac{1}{k+1}=4(k+2)+\frac{1}{k+1}>(2\sqrt{k+2})^2 $$ $$\Rightarrow$$ $$ 2\sqrt{k+2}-1< 2\sqrt{k+1}+\frac{1}{\sqrt{k+1}}-1$$ So wir have $$2\sqrt{k+2}-1< 2\sqrt{k+1}-1+\frac{1}{\sqrt{k+1}}<\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}$$

score 0 · Answer 3 · answered Jul 15 '21 at 13:39

0

Consider the curve $y=\frac 1{\sqrt x}$ and divide the interval $(0,n+1)$ into rectangles of unit length. We have $$\int_1^{n+1}\frac{1}{\sqrt{x}}\,dx\lt 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}}\lt \int_0^n \frac{1}{\sqrt{x}}\,dx$$ Evaluate the integrals, and we have tthe required proof $$2(\sqrt{n+1}-1)<1+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{n}}< 2\sqrt{n}$$

answered Jul 15 '21 at 13:39

Sayan Dutta

8,831

This is almost exactly one of the answers that are linked in the comments. Please, try to avoid rewriting the ideally same answers. To do this, take a quick look at the links that are shared in the comments claiming that the post is duplicate. – VIVID Jul 15 '21 at 13:42
@VIVID Sorry, the comment came while I was typing my answer out. That's why I missed the comment. – Sayan Dutta Jul 15 '21 at 13:45

prove that $2(\sqrt{n+1}-1)<1+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{n}}< 2\sqrt{n}$

3 Answers3