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Question Number 113465 by Algoritm last updated on 13/Sep/20

Commented by Her_Majesty last updated on 13/Sep/20

$$useyourbrains$$$$\sqrt{2019}\approx 45$$$$howcould45(\frac{1}{45+2019}+\frac{1}{45+{2019}^2}+...)$$$$begreaterthan1???$$$$ifyoustarttotrywith$$$$\sqrt{2}(\frac{1}{2+\sqrt{2}}+\frac{1}{2^2+\sqrt{2}})and\sqrt{3}(\frac{1}{3+\sqrt{3}}+...+\frac{1}{3^3+\sqrt{3}})$$$$and\sqrt{4}(\frac{1}{4+\sqrt{4}}+...+\frac{1}{4^4+\sqrt{4}})youmightsee$$$$that\sqrt{n}\underset{j=1}{\sum ^n}\frac{1}{n^i+\sqrt{n}}canonlybearational$$$$numberwhen\sqrt{n}isarationalnumber.$$$$alsoitcanneverbea\sqrt{n}witha\in \mathbb{N}$$

Commented by Algoritm last updated on 13/Sep/20

$$\mathrm{no}\mathrm{correct}$$

Commented by Her_Majesty last updated on 13/Sep/20

$$thentellme{what}^{\prime }scorrect$$

Commented by mr W last updated on 13/Sep/20

$$\mathrm{\Sigma }<2019\times \frac{\sqrt{2019}}{2019+\sqrt{2019}}=\frac{2019}{1+\sqrt{2019}}\approx 44$$$$\Rightarrow allanswersarewrong!$$

Commented by mr W last updated on 13/Sep/20

$$exactvaluefor\mathrm{\Sigma }notpossible!$$

Commented by Algoritm last updated on 13/Sep/20

$$\mathrm{answer}?$$

Commented by mr W last updated on 13/Sep/20

$$theansweristhatthequestionis$$$$wrong!ifyouhaveabetteranswer,$$$$thentellus!$$

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