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Prove that for any positive integer $n$, $$ 2\sqrt{n+1}-2 \leq 1 + 1/\sqrt{2} + 1/\sqrt{3} + .... + 1/\sqrt{n}\leq 2\sqrt{n}-1 $$

the middle part can be expressed as a sum. of $1/\sqrt{x}$ from 1 to $n$, which leads to a $Hn^{0.5}$. Im not sure how that helps.

I tried differentiating both end functions and found they are both increasing.. But it doesn't seem to help much... help pls

Simon S
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Danxe
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  • Did you try induction? – Simon S Nov 03 '14 at 02:31
  • @SimonS mathematical induction? – Danxe Nov 03 '14 at 02:32
  • Yes, mathematical induction – Simon S Nov 03 '14 at 02:33
  • SEE here; http://math.stackexchange.com/questions/1002337/prove-that-for-any-positive-integer-n-2-sqrtn1-2-le-1-frac1-sqrt2/1002342#1002342 – Paul Nov 03 '14 at 02:43
  • Instead of induction? Can i form a riemann sum with the middle portion. then deduce the f(x). use left and right rule to come up with inequality? since from the left the value will always be smaller than from the right – Danxe Nov 03 '14 at 02:56
  • Standard comparisons with Riemann integrals of the decreasing function $x\mapsto1/\sqrt{x}$ work. – Did Nov 03 '14 at 03:34

2 Answers2

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$2\sqrt{n+1}-2 \leq 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}} \leq 2\sqrt{n}-1$

We prove this by proving the following statements:

  1. $2\sqrt{n+1}-2 \leq 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}}$

  2. $1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}} \leq 2\sqrt{n}-1$

Proof of (1) by Induction:

$\\ \\$

Base case: Notice that $2\sqrt{1+1} - 2 \leq 1$.

$\\ \\$

Inductive step: Suppose $2\sqrt{k+1}-2 \leq 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{k}}$.

Then, $2\sqrt{k+1}-2 + \frac{1}{\sqrt{k+1}} \leq 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{k}} + \frac{1}{\sqrt{k+1}}$.

Thus, to show that $2\sqrt{k+2}-2 \leq 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{k}} + \frac{1}{\sqrt{k+1}}$, we can simply show that $2\sqrt{k+2}-2 \leq 2\sqrt{k+1}-2 + \frac{1}{\sqrt{k+1}}$.

\begin{align} & 2\sqrt{k+2}-2 \leq 2\sqrt{k+1}-2 + \frac{1}{\sqrt{k+1}} \\ \Longleftrightarrow & 2\sqrt{k^2+3k+2} - 2\sqrt{k+1} \leq 2k + 2 -2\sqrt{k+1} + 1 \\ \Longleftrightarrow & 2\sqrt{k^2+3k+2} \leq 2k+3 \\ \Longleftrightarrow & 4k^2+12k+8 \leq 4k^2+12k+9 \quad \text{as} \quad 2\sqrt{k^2+3k+2},2k+3 > 0 \\ \Longleftrightarrow & 8 \leq 9 \quad \text{which is true} \end{align}

Thus, (1) is proven.

Proof of (2) by Induction: $\\ \\$

Base case: Notice that $1 \leq 2\sqrt{1}-1$.

$\\ \\$

Inductive step: Suppose $1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{k}} \leq 2\sqrt{k}-1$.

Then $1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{k}} + \frac{1}{\sqrt{k+1}} \leq 2\sqrt{k}-1 + \frac{1}{\sqrt{k+1}}$.

Thus to show that $1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{k}} + \frac{1}{\sqrt{k+1}} \leq 2\sqrt{k+1}-1$, we can simply show that $2\sqrt{k}-1 + \frac{1}{\sqrt{k+1}} \leq 2\sqrt{k+1}-1$.

\begin{align} & 2\sqrt{k}-1 + \frac{1}{\sqrt{k+1}} \leq 2\sqrt{k+1}-1 \\ \Longleftrightarrow & 2\sqrt{k} + \frac{1}{\sqrt{k+1}} \leq 2\sqrt{k+1} \\ \Longleftrightarrow & 2\sqrt{k^2+k} + 1 \leq 2k+2 \\ \Longleftrightarrow & 2\sqrt{k^2+k} \leq 2k+1 \\ \Longleftrightarrow & 4k^2+4k \leq 4k^2+4k+1 \quad \text{as} \quad 2\sqrt{k^2+k},2k+1 > 0 \\ \Longleftrightarrow & 0 \leq 1 \quad \text{which is true} \end{align}

Thus, (2) is proven.

We now see that $2\sqrt{n+1}-2 \leq 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}}$ and $1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}} \leq 2\sqrt{n}-1$ are both true for all positive integers $n$. Thus, we conclude that $2\sqrt{n+1}-2 \leq 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}} \leq 2\sqrt{n}-1$ for all positive integers $n$.

  • Thanks for a clear picture!! but how come u can sub the k assumption into k+1? i'm confused with the inequalities – Danxe Nov 03 '14 at 15:29
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Hint: $2\sqrt{k+1} - 2\sqrt{k}<\dfrac{1}{\sqrt{k}} = \dfrac{2}{2\sqrt{k}} < \dfrac{2}{\sqrt{k}+\sqrt{k-1}} = 2\sqrt{k} - 2\sqrt{k-1}$. Now let $k$ runs from $1$ to $n$, and add up.

DeepSea
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  • how is it that you can just add the term sqrt(k) and sqrt(k-1) without changing the equation? – Danxe Nov 03 '14 at 02:43