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Prove an inequality of consecutive square roots using induction

by RoRi

Prove that for n \geq 1, we have

    \[ 2 \left( \sqrt{n+1} - \sqrt{n} \right) < \frac{1}{\sqrt{n}} < 2 \left( \sqrt{n} - \sqrt{n-1} \right) \]

and further that,

    \[ 2 \sqrt{m} - 2 < \sum_{n=1}^m \frac{1}{\sqrt{n}} < 2 \sqrt{m} - 1. \]

We prove each of these separately (and we use the first one in the proof of the second).
Proof #1. We prove each of the inequalities separately. First, we prove

    \[ 2 \left( \sqrt{n+1} - \sqrt{n} \right) < \frac{1}{\sqrt{n}}. \]

This just requires some algebraic manipulation,

    \begin{align*} 2 \left( \sqrt{n+1} - \sqrt{n} \right) < \frac{1}{\sqrt{n}} &\iff& 2 \sqrt{n}\sqrt{n+1} - 2n &< 1 \\ &\iff& \sqrt{n}\sqrt{n+1} &< \frac{1}{2} + n \\ &\iff& n (n+1) &< \frac{1}{4} + n + n^2 \\ &\iff& 0 &< \frac{1}{4}. \end{align*}

Since 0 < \frac{1}{4} for all n, we have the left inequality. For the right inequality we proceed similarly,

    \begin{align*} \frac{1}{\sqrt{n}} < 2\left(\sqrt{n} - \sqrt{n-1} \right) &\iff& 1 &< 2n - 2 \sqrt{n}\sqrt{n-1} \\ &\iff& \sqrt{n}\sqrt{n-1} &< n - \frac{1}{2} \\ &\iff& n(n-1) &< n^2 -n +\frac{1}{4} \\ &\iff& 0 &< \frac{1}{4}. \end{align*}

Thus, we have the inequality on the right true for all n as well. \qquad \blacksquare

Proof #2. Now, we want to use the result in the first proof to prove the second set of inequalities.
First, we consider the left inequality. For the case m=2 we have

    \[ 2\sqrt{2} - 2, \qquad \text{and} \qquad 1 + \frac{1}{\sqrt{2}}, \]

on the left and right respectively. The inequality then holds since,

    \begin{align*} 2\sqrt{2} - 2 &< 2\sqrt{2} - \sqrt{2} &(2 > \sqrt{2})\\ &= \sqrt{2} \\ &= \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} & (\sqrt{2} = \frac{2}{\sqrt{2}}) \\ &< 1 + \frac{1}{\sqrt{2}}. \end{align*}

Thus, the inequality holds in the case m =2. Assume then that the inequality holds for some k \in \mathbb{Z}_{\geq 2}. Then,

    \begin{align*} 2 \sqrt{2} - 2 < \sum_{n=1}^k \frac{1}{n} &\implies& 2\sqrt{2} - 2 + \frac{1}{\sqrt{k+1}} &< \sum_{n=1}^{k+1} \frac{1}{n} \\ &\implies& 2 \sqrt{2} - 2 + 2(\sqrt{k+2} - \sqrt{k+1}) &< \sum_{n=1}^{k+1} \frac{1}{\sqrt{n}} & (\text{Part 1})\\ &\implies& 2 \left( \sqrt{k+2} - \sqrt{k+1} \right) + 2\sqrt{k} - 2 &< \sum_{n=1}^{k+1} \frac{1}{\sqrt{n}}. \end{align*}

But, using the first part again we know 2(\sqrt{k+2} - \sqrt{k+1}) < 2(\sqrt{k+1} - \sqrt{k}); hence,

    \begin{align*} &&2(\sqrt{k+2} - \sqrt{k+1}) + 2\sqrt{k} - 2 &< \sum_{n=1}^{k+1} \frac{1}{\sqrt{n}} \\ \implies && 2\sqrt{k+1} - 2 &< \sum_{n=1}^{k+1} \frac{1}{\sqrt{n}}. \end{align*}

Thus, the inequality is true for all m \in \mathbb{Z}_{\geq 2}.

Now, for the inequality on the right. For the case m =2 we have

    \[ 1 + \frac{1}{\sqrt{2}}, \qquad \text{and} \qquad 2\sqrt{2}- 1 \]

on the left and right, respectively. But then, since \sqrt{2} = \frac{2 \sqrt{2}}{2} < \frac{3}{2} we have,

    \begin{align*} \sqrt{2} < \frac{3}{2} &\implies & 2 \sqrt{2} &< 3 \\ &\implies & 2 \sqrt{2} + 1 &< 4 \\ &\implies & 2 + \frac{1}{\sqrt{2}} &< 2 \sqrt{2} & (\text{dividing by } \sqrt{2})\\ &\implies & 1 + \frac{1}{\sqrt{2}} &< 2\sqrt{2} -1. \end{align*}

Hence, the inequality is true for the case m = 2. Assume then that it is true for some m = k \in \mathbb{Z}_{\geq 2}. Then,

    \begin{align*} \sum_{n=1}^k \frac{1}{\sqrt{n}} < 2 \sqrt{k} -1 &\implies & \left(\sum_{n=1}^k \frac{1}{\sqrt{n}} \right) + \frac{1}{\sqrt{k+1}} &< 2 \sqrt{k} - 1 + \frac{1}{\sqrt{k+1}} \\ &\implies & \sum_{n=1}^{k+1} \frac{1}{\sqrt{n}} &< 2 \sqrt{k} - 1 + 2(\sqrt{k+1} - \sqrt{k}) & (\text{Part 1})\\ &\implies & \sum_{n=1}^{k+1} \frac{1}{\sqrt{n}} &< 2 \sqrt{k+1} - 1 &(\text{Part 1, again}.) \end{align*}

Hence the right inequality holds for all m \in \mathbb{Z}_{\geq 2}.

Therefore we have established both halves of the inequality for all m \geq 2. \qquad \blacksquare

8 comments

  2. Jon Hurst says:

    I think that this proof is supposed to be about the telescoping property
    rather than induction.

    To prove:

        \[ \sum_{i=1}^{m} \frac{1}{\sqrt{i}} > 2\sqrt{m} - 2 \]

    You just need to sum the relevant inequality proved in part 1 and use
    the telescoping property:

        \[ \sum_{i=1}^{m} 2(\sqrt{i + 1} - \sqrt{i}) = 2(\sqrt{m + 1} - 1) = 2\sqrt{m + 1} - 2 \]

    and apply \sqrt{m + 1} > \sqrt{m}.

    The other inequality is slightly more complicated because

        \[ \sum_{i=1}^{m} 2(\sqrt{i} - \sqrt{n-1}) = 2\sqrt{m} \]

    and we want 2\sqrt{m} - 1. The clue is that that 2\sqrt{m} is
    valid for all m wheras 2\sqrt{m} - 1 is only valid for m \geq 2. Hence we look at:

        \[ \sum_{i=2}^{m}\frac{1}{\sqrt{i}} < \sum_{i=2}^{m}2(\sqrt{i} - \sqrt{i-1}) = \sum_{i=1}^{m}2(\sqrt{i}-\sqrt{i-1}) - 2 = 2\sqrt{m} - 2 \]

    The sum is only valid for m \geq 2 as we expected. We can then add 1
    to either side of the inequlity to find:

        \[ \sum_{i=1}^{m}\frac{1}{\sqrt{i}} < 2\sqrt{m} - 1 \]

  4. Anonymous says:

    in proof #2 , left inequality , the last step you make ,when you use the first part again ,you change small value for a bigger one but you dont know if the new left side is still going to be less than the right side.

  5. Claudio Rebelo says:

    Fist of all great work that you are doing here!
    In proof #2 there is a typo in the 1st line on the right side of the equation: a ‘2’ in the square root instead of a ‘k’.
    Everything that follows is stated false. Of course, fixing the typo makes the proof much easier ;)

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