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$$\left\lfloor \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{1024}}\right\rfloor =?$$ I can't find $$? \leq \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{n}} \leq ?$$ I remeber there was a sharp bound , Am i right ?
I am thankful if someone help me to find the bound , or give another idea
Last question : Is it possible to find $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{1024}}$ by integral ?

Darío A. Gutiérrez
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Khosrotash
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    A good approximation to use is $\int_{1}^{1024} \frac{1}{\sqrt{n}}$, which evaluates to exactly $62$. Since the area of the summation is greater than the integral (draw it out!), your summation is at least greater than $62$. Now you just have to prove the upper bound is $63$ or less. – Toby Mak Nov 16 '17 at 11:15
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    I also found another question which is quite similar here. You may want to observe the inequalities which Robert Israel uses for some inspiration. – Toby Mak Nov 16 '17 at 11:19
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    @TobyMak You said "draw it out!" How can you draw the summation? The integral can be drawn, since it is defined for individual points with width "dn". But summation elements are not defined with width "dn". Please explain. Thanks! – Gaurang Tandon Nov 16 '17 at 12:44
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    @GuarangTandon What I mean by 'drawing it out' is to find the value of the summation at that number, then draw a rectangle with that height from $x$ to $x+1$. For example, when drawing the summation for $x^2$ you will get a rectangle with height $1$ from $1$ to $2$, another one with height $4$ from $2$ to $3$, and so on. – Toby Mak Nov 16 '17 at 22:58

4 Answers4

16

Since $$\frac1{\sqrt{n+1}}<2(\sqrt{n+1}-\sqrt{n})=\frac2{\sqrt{n+1}+\sqrt{n}}<\frac1{\sqrt{n}},$$ we have $$2(\sqrt{n+1}-\sqrt{2})<\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{n}}<2(\sqrt{n}-\sqrt{1}),$$ i.e. $$2\sqrt{n+1}-2\sqrt{2}+1<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{n}}<2\sqrt{n}-1.$$ For $n=1024$, the bounds are $2\sqrt{1025}-2\sqrt{2}+1\approx62.2$ and $63$, so the result would be $62$.

5

Use

$$\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}}\in\left(\frac{1}{2\sqrt{n+1}}, \frac{1}{2\sqrt{n}}\right)$$

or, to put it differently:

$$2(\sqrt{n+1}-\sqrt{n})\lt\frac{1}{\sqrt{n}}\lt2(\sqrt{n}-\sqrt{n-1})$$

so you can squeeze your sum between two telescopic sums.

4

By the EML formula $$ H_n^{(1/2)} = 2\sqrt{n}+\zeta\left(\tfrac{1}{2}\right)+\frac{1}{2\sqrt{n}}+O\left(\frac{1}{n^{3/2}}\right) $$ hence $H_{1024}^{(1/2)}\approx \color{red}{62}+\frac{1}{2}$. There are many questions on MSE asking for approximations of $H_n^{(1/2)}$: the Hermite-Hadamard inequality/the trapezoid method for the estimation of $\int_{1}^{1024}\frac{dx}{\sqrt{x}}$ provide straightforward answers for the task at hand.

Jack D'Aurizio
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To answer your last question: no. Say we got $f(x) = 2x$, the primitive function would be $F(x) = x^2$. Now say we want to take the integral for $[0, 10]$ then we'd get $$\int_{0}^{10} f(x) dx = F(10) - F(0) = 10^2 = 100$$ However, if we'd only take integers (like in your example) the answer would be $$\sum_{x = 0}^{10}f(x) = 0 + 2 + 4 + 6 + 8 + ... + 20 = 110$$ So like, where in your question you actually got the formula $$\sum_{n=1}^{1024}\frac{1}{\sqrt{n}}$$ You can't treat it like it's the integral of $g(n) = \frac{1}{\sqrt{n}}$ because that'd give the result $$\int_{1}^{1024} g(n)dn = G(1024) - G(1) = 2 * \sqrt{1024} - 2 * \sqrt{1} = 2 * 64 - 2 = 62$$ Which does not, even though it seems like it gets very, very close, equal the summation you started with (Thanks a lot to @AlexS for correcting my mistake!). The only reason why you can't use integrals is because $dn < 1$, and, fun fact, in the past the theories about integrals were using summations, and now you accidentally tried to turn it the other way around. How funny things can be.

SuperSjoerdie
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