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this is my first post here so pardon me if I make any mistakes. I am required to prove the following, through mathematical induction or otherwise:

$$\frac{1}{\sqrt1} + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}} < 2{\sqrt{n}}$$

I tried using mathematical induction through:

$Let$ $P(n) = \frac{1}{\sqrt1} + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}} < 2{\sqrt{n}}$

$Since$ $P(1) = \frac{1}{\sqrt1} < 2{\sqrt{1}}, and$ $P(k) = \frac{1}{\sqrt1} + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{k}} < 2{\sqrt{k}},$

$P(k+1) = \frac{1}{\sqrt1} + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{k}}+ \frac{1}{\sqrt{k+1}} < 2{\sqrt{k+1}}$

Unfortunately, as I am quite new to induction, I couldn't really proceed from there. Additionally, I'm not sure how to express ${\sqrt{k+1}}$ in terms of ${\sqrt{k}}$ which would have helped me solve this question much more easily. I am also aware that this can be solved with Riemann's Sum (or at least I have seen it being solved in that way) but I do not remember nor quite understand it.

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    $\begingroup$ To be shown is that $2\sqrt{n}+\frac{1}{\sqrt{n+1}}\leq2\sqrt{n+1}$ $\endgroup$ – drhab Aug 13 '14 at 15:18
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$P(k+1)=P(k)+\dfrac{1}{\sqrt{k+1}}\leq 2\sqrt{k}+\dfrac{1}{\sqrt{k+1}}=\dfrac{2\sqrt{k}\sqrt{k+1}+1}{\sqrt{k+1}}$, in the other hand we have $2\sqrt{k}\sqrt{k+1}\leq k+k+1=2k+1$ (using $2ab\leq a^2+b^2$), now we get $2\sqrt{k}\sqrt{k+1}+1\leq 2(k+1)$, and finally $\dfrac{2\sqrt{k}\sqrt{k+1}+1}{\sqrt{k+1}}\leq 2\sqrt{k+1}$, so $P(k+1)\leq 2\sqrt{k+1}$.

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The following statements are equivalent. The first statement shows that $P(n)<2\sqrt{n}\Rightarrow P(n+1)<2\sqrt{n+1}$ and the last statement is evidently true:

$2\sqrt{n}+\frac{1}{\sqrt{n+1}}\leq2\sqrt{n+1}$

$2\sqrt{n\left(n+1\right)}+1\leq2\left(n+1\right)$

$2\sqrt{n\left(n+1\right)}\leq2n+1$

$4n\left(n+1\right)\leq4n^{2}+4n+1$

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  • $\begingroup$ Pardon me but I don't understand how $\frac{1}{\sqrt1} + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}}$ was equated to $2{\sqrt{n}}$. $\endgroup$ – calquathater Aug 14 '14 at 8:55
  • $\begingroup$ In your question $P(n)$ is a number: $\frac{1}{\sqrt{1}}+\cdots+\frac{1}{\sqrt{n}}$. In my original answer $P(n)$ was a statement: $\frac{1}{\sqrt{1}}+\cdots+\frac{1}{\sqrt{n}}<2\sqrt{n}$. This statement combined with the first statement in my answer implies statement $P(n+1)$ wich is exactly what we need to prove that $\forall n\; P\left(n\right)$ i.e. $\forall n\;\frac{1}{\sqrt{1}}+\cdots+\frac{1}{\sqrt{n}}<2\sqrt{n}$ by induction. This is my mistake. I should have mentioned this difference in my answer, and should have used another letter. I hope this makes things clear. I repaired. $\endgroup$ – drhab Aug 14 '14 at 19:41
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Add $\frac{1}{\sqrt{k+1}} $ to both sides of $P(k)$ and then show

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

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First note that $$\frac{1}{\sqrt{n}} = \frac{2}{\sqrt{n}+\sqrt{n} }\leq \frac{2}{\sqrt{n}+\sqrt{n-1}} \overset{(1)}{=} 2\left(\sqrt{n}-\sqrt{n-1} \right)$$ $(1)$ follows by multiplication with conjugate.

Now sum to get: $$\sum_ {k=1}^{n} \frac{1}{\sqrt{k}} \leq \sqrt{n}-\sqrt{0}= \sqrt{n}$$

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If you want to take a look on the Riemann's sum method :

$\forall n > 1$, we have $$\int_{n-1}^n \frac{dt}{\sqrt{t}} \ge (n-(n-1))\cdot \underset{x \in [n-1,n]}{\min} \frac{1}{\sqrt{x}} = \frac{1}{\sqrt{n}} $$

Hence, $$\frac{1}{\sqrt1} + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}} \le 1+ \int_{1}^2 \frac{dt}{\sqrt{t}} + \int_{2}^3 \frac{dt}{\sqrt{t}} + ... + \int_{n-1}^n \frac{dt}{\sqrt{t}} = 1+\int_{1}^n\frac{dt}{\sqrt{t}} $$ $$\frac{1}{\sqrt1} + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}} \le 1+\left[2\sqrt{t}\right]_1^n = 1+ (2\sqrt{n} -2 )= 2\sqrt{n} -1 < 2\sqrt{n} $$

The difference can be seen by comparing the Right Riemann Sum of the function $t \rightarrow \frac{1}{\sqrt{t}} $ to its integral on $[1,n]$.

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We have a stronger inequality. Note that $$\sqrt{x-\frac{1}{2}}+\sqrt{x+\frac{1}{2}}<2\sqrt{x}$$ for all $x\geq \dfrac{1}{2}$. Thus, for each $x>\dfrac12$, we get $$\frac{1}{\sqrt{x}}< \frac{2}{\sqrt{x-\frac12}+\sqrt{x+\frac{1}{2}}}=2\,\left(\sqrt{x+\frac12}-\sqrt{x-\frac12}\right)\,.$$ That is, for every nonnegative integer $n$, $$\sum_{k=1}^n\,\frac{1}{\sqrt{k}}\leq 2\,\sum_{k=1}^n\,\left(\sqrt{k+\frac12}-\sqrt{k-\frac12}\right)=2\,\left(\sqrt{n+\frac12}-\sqrt{\frac12}\right)\,.$$ The equality holds if and only if $n=0$. This is a sharper inequality than the required inequality as $$\sqrt{x+y}\leq \sqrt{x}+\sqrt{y}$$ for any $x,y\geq 0$ (with equality iff $x=0$ or $y=0$).

On the other hand, you can show that $$2\sqrt{x}\leq \sqrt{x-\frac{7}{16}}+\sqrt{x+\frac{9}{16}}$$ for every $x\geq 1$ (with equality case $x=1$). This gives $$\sum_{k=1}^n\,\frac{1}{\sqrt{k}}\geq 2\,\left(\sqrt{n+\frac{9}{16}}-\frac{3}{4}\right)$$ for all integers $n\geq 0$ with equality cases $n=0$ and $n=1$. That is, $$2\,(\sqrt{n+1}-1)\leq 2\,\left(\sqrt{n+\frac{9}{16}}-\frac{3}{4}\right)\leq \sum_{k=1}^n\,\frac{1}{\sqrt{k}}\leq 2\,\left(\sqrt{n+\frac12}-\sqrt{\frac12}\right)\leq 2\sqrt{n}$$ for every nonnegative integer $n$. If $n=0$, then every inequality is an equality. If $n=1$, then only the second inequality from the left becomes an inequality. For $n>1$, all inequalities are strict.

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