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We have to find the floor $\lfloor S \rfloor$ of the following sum:

$$S = \sum_{k=1}^{80}\frac{1}{\sqrt k}$$

What I did was to find a approximate series that this series is near to. Let that series have general term $T_k$ and original series may have general term $a_k$. We construct the following series of $T_k$

$$T_k = \frac{1}{ \sqrt{k+1}+\sqrt{k}} = \sqrt{k+1}-\sqrt{k}\\$$

Then we have the following inequality:

$$\frac{a_k}{2} = \frac{1}{\sqrt{k}+\sqrt{k}} > T_k \\ \sum a_k > 2 \sum_{1}^{80} T_k \\ S > 2 (\sqrt{81}-1)$$

Where the last result is due to telescoping property of $T_k$. So we have a lower limit $ S_k >\color{indigo}{ 16}$

However still we cannot say $\lfloor S \rfloor = 16$ because $S$ may exceed $17$.

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    $\begingroup$ Find the sum going to $k=81$. That's your upper bound. $\endgroup$ – stuart stevenson Dec 17 '17 at 18:31
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    $\begingroup$ @Steve Thanks but I don't think it leads me anywhere! $\endgroup$ – samjoe Dec 17 '17 at 18:37
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We have that $\sqrt{n+a+1}-\sqrt{n+a}$ behaves like $\frac{1}{2\sqrt{n}}$ for large values of $n$, and by picking $a=-\frac{1}{2}$ we get a telescopic term providing an accurate approximations of $\frac{1}{\sqrt{n}}$ for any $n\geq 1$: $$ 2\sqrt{n+\tfrac{1}{2}}-2\sqrt{n-\tfrac{1}{2}} =\frac{1}{\sqrt{n}}+E(n),\qquad \frac{1}{32 n^2\sqrt{n}}\leq E(n)\leq \frac{1}{28n^2\sqrt{n}} $$ It follows that $$ \sum_{n=1}^{80}\frac{1}{\sqrt{n}} = 2\sqrt{80+\tfrac{1}{2}}-2\sqrt{\tfrac{1}{2}}+\theta,\qquad |\theta|\leq\frac{1}{28}\zeta\left(\frac{5}{2}\right) $$ hence $\sum_{n=1}^{80}\frac{1}{\sqrt{n}}$ belongs for sure to the interval $(16,17)$ and it is pretty close to the midpoint of such interval.

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    $\begingroup$ Thanks but you said it's valid for large n only? $\endgroup$ – samjoe Dec 18 '17 at 4:01
  • $\begingroup$ @samjoe: for any $a>0$, $\sqrt{n+a+1}-\sqrt{n+1}$ behaves like $\frac{1}{2\sqrt{n}}$ for large values of $n$. If we carefully pick $a$ as $-\frac{1}{2}$, the shown inequalities hold for any $n\geq 1$. $\endgroup$ – Jack D'Aurizio Dec 18 '17 at 4:04
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    $\begingroup$ Thanks a lot ! Only question is how you got bounds of the error term is it binomial expansion $\endgroup$ – samjoe Dec 18 '17 at 4:13
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    $\begingroup$ @samjoe: with a bit of patience, you can rationalize $2\sqrt{n+1/2}-2\sqrt{n-1/2}-\frac{1}{\sqrt{n}}$ and check that such function multiplied by $n^2\sqrt{n}$ is bounded and decreasing (from something a bit smaller than $\frac{1}{28}$ to $\frac{1}{32}$). $\endgroup$ – Jack D'Aurizio Dec 18 '17 at 4:15
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$$\begin{align} \int_1^{81}\frac 1{\sqrt x}\;\;\text d x &<\qquad\sum_{k=1}^{80}\frac 1{\sqrt k} &&<1+\int_1^{80}\frac 1{\sqrt x}\;\;\text d x\\ \bigg[2\sqrt{x}\bigg]_1^{81} &< \qquad\sum_{k=1}^{80}\frac 1{\sqrt k} &&<1+\bigg[2\sqrt{x}\bigg]_1^{80}\\ 2\big(\sqrt {81}-\sqrt{1}\big) &< \qquad\sum_{k=1}^{80}\frac 1{\sqrt k} &&< 1+2\big(\sqrt{80}-\sqrt{1}\big)\\ 16 &< \qquad\sum_{k=1}^{80}\frac 1{\sqrt k} &&<16.88 \end{align}$$

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NB: Wolframalpha gives $16.484$.

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  • $\begingroup$ how did you get the first inequality? $\endgroup$ – Abcd Jan 11 at 13:52
  • $\begingroup$ @Abcd - by considering the graph of $\frac 1{\sqrt{k}}$. $\endgroup$ – hypergeometric Jan 11 at 14:16
  • $\begingroup$ Not sure how the sum gets sandwhiched between the integrals. I feel it should have the least value because its like a subset of the two integrals. $\endgroup$ – Abcd Jan 11 at 14:20
  • $\begingroup$ Just draw the graph and it will become clear!. The integral is the area under the curve and the summation is the sum of areas of rectangular bars of unit width. $\endgroup$ – hypergeometric Jan 11 at 14:22
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You can use Abel Sum:

we get $$S\left(N\right)=\sum_{k=1}^{N}\frac{1}{\sqrt{k}}=\sqrt{N}+\frac{1}{2}\int_{1}^{N}\frac{\left\lfloor t\right\rfloor }{t^{3/2}}dt $$

so, using the bounds

$$t-1\leq\left\lfloor t\right\rfloor \leq t $$

we have

$$2\sqrt{N}+\frac{1}{\sqrt{N}}-2\leq S(N)\leq2\sqrt{N}-1 $$

hence

$$\left\lfloor S\left(80\right)\right\rfloor =16.$$

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    $\begingroup$ Omg what is this. Thank you but we haven't covered this yet. Could you please explain? Thanks alot! $\endgroup$ – samjoe Dec 17 '17 at 18:37
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    $\begingroup$ This is a sort of mathematical trick, called Abel summation: en.wikipedia.org/wiki/Abel%27s_summation_formula The rest is just a boundary manipulation! Abel summation can help a lot! $\endgroup$ – Von Neumann Dec 17 '17 at 18:40
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    $\begingroup$ @HenryTuring With the notation used in Wikipedia, what are here $a_n$ and $\phi(x)$? $\endgroup$ – ajotatxe Dec 17 '17 at 18:49
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    $\begingroup$ @ajotatxe $a_n = 1$ and $\phi(x) = \frac{1}{\sqrt{x}}$ $\endgroup$ – Von Neumann Dec 17 '17 at 19:57
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If you now use the fact that $\tfrac{a_k}2<T_{k-1}$ (because $\frac1{2\sqrt k}<\frac1{\sqrt k +\sqrt{k-1}}=T_{k-1}$) you'll find that the sum has to be strictly smaller than $2\sqrt{80}$, so you now know that the answer is either $16$ or $17$.

But a little more detail shows that $$\sum_{k=2}^{80}\frac1{\sqrt{k}}<2\sum_{k=2}^{80}\left(\sqrt k - \sqrt{k-1}\right)=2(\sqrt{80}-1)<16;$$ but then $$\sum_{k=1}^{80}\frac1{\sqrt{k}}=\frac1{\sqrt 1}+\sum_{k=2}^{80}\frac1{\sqrt{k}}<1+16=17$$ (note that this is strict; actually $2(\sqrt{80}-1)+1\simeq16.889$). So the result is $16$.

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  • $\begingroup$ Thanks for this answer! I wonder why I didn't try this out.. $\endgroup$ – samjoe Dec 18 '17 at 3:54

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