Induction: Bounds on sum of inverses of first $n$ square roots

Question

I am trying to show by induction that $2(\sqrt{n}-1)<1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}<2\sqrt{n}$.
For the upper bound, I have that if $1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}<2\sqrt{n}$, then $1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n+1}}<2\sqrt{n}+\dfrac{1}{\sqrt{n+1}}=$

$2\sqrt{n}+\dfrac{2}{2\sqrt{n+1}}<2\sqrt{n}+\dfrac{2}{\sqrt{n}+\sqrt{n+1}}=2\sqrt{n}+2(\sqrt{n+1}-\sqrt{n})=2\sqrt{n+1}$.

However, despite hours of trying, I cannot figure out how to show that $2(\sqrt{n+1}-1)<1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n+1}}$.

Answer 1

Observe that: $\dfrac{1}{2\sqrt{k}}> \dfrac{1}{\sqrt{k+1}+\sqrt{k}}= \sqrt{k+1}-\sqrt{k}$. Taking summation for $k$ from $1$ to $n$ and get: $\dfrac{S_n}{2} > \displaystyle \sum_{k=1}^n\left(\sqrt{k+1} - \sqrt{k}\right)= \sqrt{n+1}-1\implies S_n > 2\left(\sqrt{n+1}-1\right)> 2\left(\sqrt{n}-1\right)$ as claimed.

Answer 2

$$\begin{array}{cl} & 1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}} \\ >& \dfrac2{1+\sqrt2}+\dfrac{2}{\sqrt{2}+\sqrt3}+\cdots+\dfrac{2}{\sqrt{n}+\sqrt{n+1}} \\ =& \dfrac{2(\sqrt2-1)}{(\sqrt2+1)(\sqrt2-1)}+\dfrac{2(\sqrt3-\sqrt2)}{(\sqrt{3}+\sqrt2)(\sqrt3-\sqrt2)}+\cdots+\dfrac{2(\sqrt{n+1}-\sqrt n)}{(\sqrt{n+1}+\sqrt{n})(\sqrt{n+1}-\sqrt n)} \\ =& 2(\sqrt2-1)+2(\sqrt3-\sqrt2)+\cdots+2(\sqrt{n+1}-\sqrt n) \\ =& 2\sqrt{n+1}-2 \\ \end{array}$$

Answer 3

Try to use the induction method for $2(\sqrt{n}-1+\frac{1}{\sqrt{n}}) < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}}$

Check it for base cases like $n=1,2,3,4,5,6,7,8,9,10$ and it will fail for all $n=1,2,3,4,5,6,7$ and will hold for $n=8,9,10$.

Induction step : assume that $2(\sqrt{n}-1+\frac{1}{\sqrt{n}}) < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}}$ is true for the number $n$.

Proving step : we need to prove that $2(\sqrt{n+1}-1+\frac{1}{\sqrt{n+1}}) < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}$

From the induction step we know that $2(\sqrt{n}-1+\frac{1}{\sqrt{n}}) < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}}$ thus replacing $ \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}}$ with $2(\sqrt{n}-1+\frac{1}{\sqrt{n}})$ will strengthen the inequality.

We arrive at $2(\sqrt{n+1}-1+\frac{1}{\sqrt{n+1}}) < 2(\sqrt{n}-1+\frac{1}{\sqrt{n}})+\frac{1}{\sqrt{n+1}}$, (i will leave this for you to prove).

Now since its obvious that $2(\sqrt{n}-1+\frac{1}{\sqrt{n}}) > 2(\sqrt{n}-1)$ then the inequality $2(\sqrt{n}-1) < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}}$ will be a corollary to this induction proof.

you just have to check basic cases for $2(\sqrt{n}-1) < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}}$ which are $n=1,2,3,4,5,6,7$ and thus completing the proof.

Answer 4

Note: I can see that there are already fine answers using induction!

An ad in favor of graph paper. The outcome of the standard sum/integral comparison is $$ 2 \sqrt {n+1} - 2 \; < \; 1 + \frac{1}{\sqrt 2 } + \frac{1}{\sqrt 3 } + \cdots + \frac{1}{\sqrt n } \; < \; 2 \sqrt n - 1 $$