Apostol's Calculus, exercise number I 4.7 13.

Prove that if $n \geq 1$, then $$ 2(\sqrt{n+1} - \sqrt{n}) < \frac{1}{\sqrt{n}} < 2(\sqrt{n} - \sqrt{n-1}) $$ and use this to prove that if $m \geq 2$, then $$ 2 \sqrt{m} - 2 < \sum_{n = 1}^m \frac{1}{\sqrt{n}} < 2\sqrt{m} - 1 $$ In particular, when $m = 10^6$ the sum lies between $1998$ and $1999$.

After establishing the trivial case, for $n=1$, I can't think of a way to perform the inductive step:

assuming $2(\sqrt{n+1} - \sqrt{n}) < \frac{1}{\sqrt{n}} < 2(\sqrt{n} - \sqrt{n-1})$ and using this, deducing the same for $n+1$ is true as well:

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

Any help is very appreciated, thanks

  • $\begingroup$ You should at least tell us what you tried to solve this and where are you having problems, exactly. Also, please note that we prefer to avoid questions in picture form here. Please have a look at our basic MathJax tutorial to learn how to write math here. $\endgroup$
    – A.P.
    Mar 29, 2015 at 20:51
  • $\begingroup$ You really have shown no effort whatsoever. In fact, it is not even clear if you're having a problem with the first task, with the second task, or with both. Improve your question and show some effort attempting to answer it on your own if you're expecting others to make the same effort. $\endgroup$ Mar 29, 2015 at 20:51
  • $\begingroup$ I'm sorry, it's my first post! $\endgroup$
    – Bog
    Mar 29, 2015 at 20:53
  • $\begingroup$ I'm having trouble proving the first part by induction, I'd like to know how to do it in order to proceed to the second part. I'm definitely going to learn how to write math properly here on stackexchange, sorry again $\endgroup$
    – Bog
    Mar 29, 2015 at 20:59

1 Answer 1


Method 1: (the easiest one)

$$\sqrt{x+1}-\sqrt{x}=\dfrac1{\sqrt{x+1}+\sqrt{x}},\tag1$$ so your inequality is equivalent to : $$ \begin{align*} &\dfrac2{\sqrt{n+1}+\sqrt{n}}\lt\dfrac1{\sqrt{n}}\lt\dfrac2{\sqrt{n}+\sqrt{n-1}}\tag2\\&\iff\dfrac1{\sqrt{n+1}+\sqrt{n}}\lt\dfrac1{2\sqrt{n}}\lt\dfrac1{\sqrt{n}+\sqrt{n-1}}\tag3\\&\iff\sqrt{n}+\sqrt{n-1}\lt2\sqrt{n}\lt\sqrt{n+1}+\sqrt{n}\tag4\\&\iff\sqrt{n-1}\lt\sqrt{n}\lt\sqrt{n+1}.\tag5 \end{align*}$$ This is the crucial part, we have : $$ \sqrt{n-1}\lt\color{#C00}{\sqrt{n}\lt\sqrt{n+1}\lt\sqrt{n+2}}.\tag6 $$ Adding to each side $\sqrt{n+1}$ one obtains : $$ \sqrt{n}+\sqrt{n+1}\lt2\sqrt{n+1}\lt\sqrt{n+1}+\sqrt{n+2}.\tag7 $$ Taking the reciprocal of each side and then multiplying by $2$ yields : $$ \dfrac2{\sqrt{n+1}+\sqrt{n+2}}\lt\dfrac1{\sqrt{n+1}}\lt\dfrac2{\sqrt{n}+\sqrt{n+1}}.\tag8 $$

$(2)$ is $P(n)$ and $(8)$ is $P(n+1)$. We showed that $P(n)\implies P(n+1)$. The base case holds. Hence by induction $(2)$ holds for all $n\geqslant1$. $$\tag*{$\tiny\blacksquare$}$$

Method 2:

By multiplying each side of $(4)$ by $\tfrac{\sqrt{n+1}}{\sqrt{n}}$ one gets : $$ \sqrt{n+1}+\dfrac{\sqrt{n+1}\sqrt{n-1}}{\sqrt{n}}\lt2\sqrt{n+1}\lt\sqrt{n+1}+\dfrac{n+1}{\sqrt{n}}.\tag9 $$ Note: This is the same as the $$\dfrac1{\color{red}{\frac{n+1}{\sqrt{n}}}+\sqrt{n+1}}\lt\frac1{2\sqrt{n+1}}\lt‌​\dfrac1{\sqrt{n}+\color{red}{\frac{\sqrt{n+1}\sqrt{n-1}}{\sqrt{n}}}},$$ I referred to in the comment section. Anyway, to prove $(9)$ we first show that: $$\begin{align*} \sqrt{n+1}+\dfrac{\sqrt{n+1}\sqrt{n-1}}{\sqrt{n}}\lt2\sqrt{n+1}&\iff\dfrac{\sqrt{n+1}\sqrt{n-1}}{\sqrt{n}}\lt\sqrt{n+1}\quad\tag{10}\\ &\iff\dfrac{\sqrt{n-1}}{\sqrt{n}}\lt1\tag{11}\\ &\iff\sqrt{n-1}\lt\sqrt{n}.\tag{12} \end{align*}$$ $(12)$ is obviously true hence $(10)$ holds.

Now moving on to showing the second part : $$\begin{align*} 2\sqrt{n+1}\lt\sqrt{n+1}+\dfrac{n+1}{\sqrt{n}}&\iff\sqrt{n+1}\lt\dfrac{n+1}{\sqrt{n}}=\dfrac{\sqrt{n+1}^2}{\sqrt{n}}\tag{13}\\ &\iff1\lt\dfrac{\sqrt{n+1}}{\sqrt{n}}\tag{14}\\ &\iff\sqrt{n}\lt\sqrt{n+1}\tag{15}.\\ \end{align*}$$

$(15)$ is obviously true hence $(13)$ holds. So by this last result and the former we conclude that $(9)$ holds. $$\tag*{$\tiny\blacksquare$}$$

  • $\begingroup$ how can I prove it by using induction? I mean, after establishing the first step as n=1 how can I prove the same inequalities with n+1 starting from those I posted (with n) assumed true? $\endgroup$
    – Bog
    Mar 29, 2015 at 21:11
  • $\begingroup$ @Bog What do you need to move from $$\dfrac1{\sqrt{n+1}+\sqrt{n}}\lt\dfrac1{2\sqrt{n}}\lt\dfrac1{\sqrt{n}+\sqrt{n-1}},\tag*{$P(n)$}$$ to $$\dfrac1{\sqrt{n+2}+\sqrt{n+1}}\lt\dfrac1{2\sqrt{n+1}}\lt\dfrac1{\sqrt{n+1}+ \sqrt{n}}?\tag*{$P(n+1)$}$$ $\endgroup$
    – Workaholic
    Mar 30, 2015 at 11:07
  • $\begingroup$ if I knew I wouldn't ask.. $\endgroup$
    – Bog
    Mar 30, 2015 at 11:13
  • $\begingroup$ @Bog Try to multiply the first inequality by $\frac{\sqrt{n}}{\sqrt{n+1}}$. $\endgroup$
    – Workaholic
    Mar 30, 2015 at 11:23
  • $\begingroup$ I already tried what you suggest, and other algebraic manipulations too, but I get stuck and can't get to the desired conclusion.. $\endgroup$
    – Bog
    Mar 30, 2015 at 11:26

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