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If $\displaystyle A=\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+.........+\frac{1}{100\sqrt{99}}\;,$ Then $\lfloor A \rfloor =$

Where $\lfloor x \rfloor$ represent floor function of $x$.

$\bf{My\; Try:}$ For lower bound $$\sum^{99}_{k=1}\frac{1}{(k+1)\sqrt{k}}>\sum^{99}_{k=1}\frac{1}{(k+1)k}=\sum^{99}_{k=1}\left[\frac{1}{k}-\frac{1}{k+1}\right]=1-\frac{1}{99}$$

Now I didn't understand how can I solve it, any help?

Thanks

Community
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juantheron
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    I think your lower bound should be $1-\frac 1{100}$... You might consider using bounds $1\over k^\frac 32$ and $1\over (k+1)^\frac32$ – abiessu Jan 31 '16 at 15:56
  • To abiessu actually answer given as $\displaystyle 1<A<2$, I did not understand how we can use $\displaystyle \frac{1}{k\sqrt{k}}$ and $\displaystyle \frac{1}{(k+1)\sqrt{k+1}}$ – juantheron Jan 31 '16 at 15:59
  • My thought was that we could use an integral approximation from there... – abiessu Jan 31 '16 at 16:04
  • Where did you get that question ? – Sudhanshu Jan 31 '16 at 16:07

4 Answers4

8

We have: $$ \frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}} = \frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k(k+1)}}=\frac{1}{k\sqrt{k+1}+(k+1)\sqrt{k}}\geq \frac{1}{2(k+1)\sqrt{k}}$$ hence: $$ \sum_{k=1}^{99}\frac{1}{(k+1)\sqrt{k}}\leq 2\sum_{k=1}^{99}\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)=2-\frac{1}{5}$$ but in a similar way: $$ \sum_{k=1}^{99}\frac{1}{(k+1)\sqrt{k}}\geq \frac{1}{2}+2\sum_{k=2}^{99}\left(\frac{1}{\sqrt{k-1/4}}-\frac{1}{\sqrt{k+3/4}}\right)=\frac{1}{2}+\frac{4}{\sqrt{7}}-\frac{4}{\sqrt{399}}$$ so $\lfloor A \rfloor = \color{red}{1}$. Another great victory for creative telescoping.

Jack D'Aurizio
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4

This series is a Riemann sum for $$\int_1^{99} \frac {dx}{(1+x)\sqrt x}$$

But that integral can be computed "exactly", as the indefinite integral is $2\,tan^{-1}(\sqrt x)$. We get about $1.37$ for the definite integral, and it's not hard to see that this makes your answer $1$.

lulu
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4

Let

$$A = \sum_{k=1}^{99} \frac1{(k+1) \sqrt{k}} $$

You can show that the sum is greater than $1$. First we can lower bound the sum by

$$\sum_{k=1}^{99} \frac1{k (k+1)} = \frac{99}{100} $$

which, while less than one, may be corrected by its second term:

$$\frac{99}{100} + \frac13 \left (\frac1{\sqrt{2}} - \frac12 \right ) \gt 1$$

i.e., $A \gt 1$. Next, consider that

$$A = \sum_{k=1}^{99} \frac1{(k+1) \sqrt{k}} = \sum_{k=1}^{99} \left (\frac{\sqrt{k}}{k} - \frac{\sqrt{k+1}}{k+1} \right ) + \sum_{k=1}^{99} \frac{\sqrt{k+1}-\sqrt{k}}{k+1}$$

The first sum on the RHS is just $9/10$. The second sum, however, is

$$ \sum_{k=1}^{99} \frac1{(k+1) (\sqrt{k+1}+\sqrt{k})} \le \frac12 \sum_{k=1}^{99} \frac1{(k+1) \sqrt{k}} = \frac12 A$$

Thus $A \le \frac9{10} + \frac12 A$, or

$$A = \sum_{k=1}^{99} \frac1{(k+1) \sqrt{k}} \le \frac95$$

and is also greater than one. Thus, the floor of the sum is $1$.

Ron Gordon
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3

Hint. Observe that, for $x\in [k,k+1],\, k=1,2,3,\cdots$, $$ \frac1{(k+2)\sqrt{k+1}} \leq\frac1{(x+1)\sqrt{x}}\leq \frac1{(k+1)\sqrt{k}}. $$ Integrating both sides from $x=k$ to $x=k+1$ gives $$ \frac1{(k+2)\sqrt{k+1}} \leq\int_k^{k+1}\frac1{(x+1)\sqrt{x}}dx\leq \frac1{(k+1)\sqrt{k}} $$ then, summing from $k=1$ to $k=99$, $$ \sum_{k=1}^{100}\frac1{(k+1)\sqrt{k}}-\frac12\leq\int_1^{100}\frac1{(x+1)\sqrt{x}}dx\leq \sum_{k=1}^{99}\frac1{(k+1)\sqrt{k}}. $$ We have $$ \int_1^{100}\frac1{(x+1)\sqrt{x}}dx=2\arctan (10) -\frac{\pi}2. $$

It follows that

$$ 2\arctan (10) -\frac{\pi}2\leq\sum_{k=1}^{99}\frac1{(k+1)\sqrt{k}}\leq2\arctan (10) -\frac{\pi}2+\frac12-\frac1{1010}. $$ or $$ \color{red}{1.371\cdots}<\sum_{k=1}^{99}\frac1{(k+1)\sqrt{k}}<\color{blue}{1.870\cdots}. $$

Olivier Oloa
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