Prove $1 + 1/2 + 1/3 + \ldots + 1/n < \sqrt{n}$ for $n \ge 7$

Question

I was given a quiz in class today and there was a question a that looked like this

Prove 1 + $\frac{1}{2}$ + $\frac{1}{3}$ + ... + $\frac{1}{n}$ < $\sqrt{n}$

for { $n\geq 7$ }

I was asked to prove this inequality using mathematical induction. The professor will not go over the question in class. I was wondering if i was showing the right steps to proving this and if someone could finish it for me or show alternative way of doing it. This is what i have some far.

step #1 :Test for n = 7

$\frac{1}{n}$ < $\sqrt{n}$

1/7 < $\sqrt{7}$

statement true for n = 7

step #2 : Induction hypothesis

Assume statement is true for some n = k

1 + $\frac{1}{2}$ + $\frac{1}{3}$ + ... + $\frac{1}{k}$< $\sqrt{k}$

for { $k\geq 7$ }

step #3 : Induction step

prove statement is true for all n = k +1

1 + $\frac{1}{2}$ + $\frac{1}{3}$ + ... + $\frac{1}{k}$ + $\frac{1}{k + 1}$ < $\sqrt{k + 1}$

assume 1 + $\frac{1}{2}$ + $\frac{1}{3}$ + ... + $\frac{1}{k}$< $\sqrt{k}$

for { $k\geq 7$ }

1 + $\frac{1}{2}$ + $\frac{1}{3}$ + ... + $\frac{1}{k}$ + $\frac{1}{k + 1}$ < $\sqrt{k}$ + $\frac{1}{k + 1}$

prove $\sqrt{k}$ + $\frac{1}{k + 1}$ < $\sqrt{k + 1}$

Answer 1

It is known that $$\lim_{n\rightarrow\infty}H_n-\ln (n+1)=\gamma$$Where $\gamma$ is the Euler-Mascheroni constant and $H_n$ is the sum in the OP. To prove that $\gamma$ exists, you could rewrite $\ln (n+1)$ as $$\ln (n+1)=\sum_{k=1}^n\ln\left(1+\frac1k\right)$$By telescoping. Then the limit is equivalent to the infinite series $$\sum_{k=1}^\infty\left(\frac1k-\ln\left(1+\frac1k\right)\right)$$Proving this converges is simple by the integral test. Now, what does this have to do with the question? Well, we just showed that $H_n<\ln(n+1)+\gamma$. If we could prove that $\ln(n+1)+\gamma<\sqrt{n}$ at some point then we have proved the claim. All you need to do is take the ratio of these two functions and show it approaches $0$ by L'Hopital's rule. And then we are done.

Answer 2

Your initial step doesn't make sense. For $n = 7$, the inequality to be proven is $$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} < \sqrt{7}.$$ It's not just $\frac{1}{7} < \sqrt{7}$.

To prove the induction step, you need to show $$\sqrt{n} + \frac{1}{n+1} < \sqrt{n+1}.$$ Hint: observe that $$\sqrt{n+1} - \sqrt{n} = \frac{(\sqrt{n+1} - \sqrt{n})(\sqrt{n+1} + \sqrt{n})}{\sqrt{n+1} + \sqrt{n}}.$$

Answer 3

Dovetailing into the proof by induction requirement:
alternative method of proving that for $~n \geq 7,~$
$\displaystyle \frac{1}{n+1} < \sqrt{n+1} - \sqrt{n}.~$

Let $~\displaystyle I = \int_n^{n+1} \frac{1}{x}\,dx.~$
Since the function $~\dfrac{1}{x}~$ is strictly decreasing, $~\dfrac{1}{n+1} < I.$

Let $~\displaystyle J = \frac{2}{3} \int_n^{n+1} x^{3/2} \,dx \implies \sqrt{n+1} - \sqrt{n} = J.$

Further, for $~\displaystyle x > 2, ~ x^{5/2} > \frac{3}{2} \implies \frac{2}{3} \,x^{3/2} > \frac{1}{x}.$

Therefore,

$$\sqrt{n+1} - \sqrt{n} = J > I > \frac{1}{n+1}.$$