Question Number 67651 by ~ À ® @ 237 ~ last updated on 29/Aug/19
![1)Let consider S= Σ_(n=0) ^∞ n and T=Σ_(n=0) ^∞ (−1)^(n+1) n We know that ∀ x∈]−1;1] Σ_(n=0) ^∞ (−x)^n =(1/(1+x)) , then after derivating (1/((1+x)^2 ))=Σ_(n=1) ^∞ (−1)^(n+1) nx^(n−1) for x=1 ,we get T=(1/4) Now let ascertain something T=Σ_(n=0) ^∞ (−1)^(2n+2) (2n+1) +Σ_(n=0) ^∞ (−1)^(2n+1) (2n) =Σ_(n=0) ^∞ (2n+1) −2S (•) knowing that S=Σ_(n=0) ^∞ (2n)+Σ_(n=0) ^∞ (2n+1) So Σ_(n=0) ^∞ (2n+1)= S−2S When replacing that value in (•) we get T=(S−2S )−2S If i conclude that T=−3S and finally find S=−(T/3)=−(1/(12)) where will the mistake be? 2)Let consider K=1+((2020)/(2019))+(((2020)/(2019)))^2 +(((2020)/(2019)))^3 +......+... ((2020)/(2019))K=((2020)/(2019))+(((2020)/(2019)))^2 +(((2020)/(2019)))^3 +.... K−1=((2020)/(2019))+(((2020)/(2019)))^2 +(((2020)/(2019)))^3 +... So ((2020)/(2019))K=K−1 Then K=−2019 Where is the error? 3)Let consider n an integer We have 0=n−n =n+(−n)=n+(−n)^(2×(1/2)) =n+[(−n)^2 ]^(1/2) =n+ (√n^2 ) = n+n=2n So << all integer are null : 0 is the only integer>> Where is the error? 4) let consider n an integer different of zero and f(n)=nln(n) we have (df/dn)=ln(n)+1 (•) Likewise f(n)=ln(n^n ) and we know that n^n =n×n×n×......×n (n times) So f(n)=ln(n)+ln(n)+......+ln(n) (n times) Now we have (df/dn)=(1/n)+(1/n)+....+(1/n) (n times) So (df/( dn))=1 (••) Relation (•) and (••) give ln(n)+1=1 then ln(n)=0 ⇒ n=1 << The logarithm of all n≥1 is null : There is no integer big than 1 >> Where is the error? 5) let consider x=0,999999999....... we ascertain that 10x=9,999999999...... then 10x=9+0,999999999.... So 10x=9+x finally x=1 << 0.9999999999999999 ..... is and integer >> Is there any error? 6)let consider a=((26666666666666666)/(66666666666666665)) b=((999999999999999999999995)/(199999999999999999999999)) In the way to cancel , if i just remove one common figure to the numerator and to the denominator And i find a=(2/5) and b=(5/1) Will it be wrong? if no ,explain!](https://www.tinkutara.com/question/Q67651.png)
$$\left.\mathrm{1}\right){Let}\:{consider}\:\:{S}=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:{n}\:\:\:\:\:{and}\:\:{T}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {n} \\ $$$$\left.{W}\left.{e}\:{know}\:{that}\:\:\forall\:{x}\in\right]−\mathrm{1};\mathrm{1}\right] \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−{x}\right)^{{n}} \:=\frac{\mathrm{1}}{\mathrm{1}+{x}}\:,\:\: \\ $$$${then}\:{after}\:{derivating}\: \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \:{nx}^{{n}−\mathrm{1}} \\ $$$${for}\:\:{x}=\mathrm{1}\:,{we}\:{get}\:\:\:{T}=\frac{\mathrm{1}}{\mathrm{4}}\: \\ $$$${Now}\:\:{let}\:{ascertain}\:{something} \\ $$$${T}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{2}{n}+\mathrm{2}} \left(\mathrm{2}{n}+\mathrm{1}\right)\:+\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{2}{n}+\mathrm{1}} \left(\mathrm{2}{n}\right) \\ $$$$\:\:=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\mathrm{2}{n}+\mathrm{1}\right)\:−\mathrm{2}{S}\:\:\:\:\:\:\:\:\:\:\:\:\left(\bullet\right) \\ $$$${knowing}\:{that}\:{S}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\mathrm{2}{n}\right)+\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\mathrm{2}{n}+\mathrm{1}\right) \\ $$$${So}\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\mathrm{2}{n}+\mathrm{1}\right)=\:{S}−\mathrm{2}{S}\: \\ $$$${When}\:{replacing}\:{that}\:{value}\:{in}\:\:\left(\bullet\right) \\ $$$${we}\:{get}\:\: \\ $$$${T}=\left({S}−\mathrm{2}{S}\:\right)−\mathrm{2}{S} \\ $$$$\: \\ $$$${If}\:{i}\:{conclude}\:{that}\:\:{T}=−\mathrm{3}{S}\:\: \\ $$$${and}\:{finally}\:\:{find}\:\:{S}=−\frac{{T}}{\mathrm{3}}=−\frac{\mathrm{1}}{\mathrm{12}}\:\: \\ $$$${where}\:{will}\:{the}\:{mistake}\:{be}? \\ $$$$\left.\:\mathrm{2}\right){Let}\:{consider}\:{K}=\mathrm{1}+\frac{\mathrm{2020}}{\mathrm{2019}}+\left(\frac{\mathrm{2020}}{\mathrm{2019}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2020}}{\mathrm{2019}}\right)^{\mathrm{3}} +……+… \\ $$$$\frac{\mathrm{2020}}{\mathrm{2019}}{K}=\frac{\mathrm{2020}}{\mathrm{2019}}+\left(\frac{\mathrm{2020}}{\mathrm{2019}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2020}}{\mathrm{2019}}\right)^{\mathrm{3}} +…. \\ $$$${K}−\mathrm{1}=\frac{\mathrm{2020}}{\mathrm{2019}}+\left(\frac{\mathrm{2020}}{\mathrm{2019}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2020}}{\mathrm{2019}}\right)^{\mathrm{3}} +… \\ $$$${So}\:\:\frac{\mathrm{2020}}{\mathrm{2019}}{K}={K}−\mathrm{1}\: \\ $$$${Then}\:\:{K}=−\mathrm{2019} \\ $$$${Where}\:{is}\:{the}\:{error}? \\ $$$$\left.\mathrm{3}\right){Let}\:{consider}\:{n}\:{an}\:\:{integer}\: \\ $$$${We}\:{have} \\ $$$$\mathrm{0}={n}−{n} \\ $$$$\:\:={n}+\left(−{n}\right)={n}+\left(−{n}\right)^{\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\:={n}+\left[\left(−{n}\right)^{\mathrm{2}} \right]^{\frac{\mathrm{1}}{\mathrm{2}}} ={n}+\:\sqrt{{n}^{\mathrm{2}} }\:=\:{n}+{n}=\mathrm{2}{n} \\ $$$${So}\:<<\:{all}\:\:{integer}\:\:\:{are}\:{null}\:\::\:\:\mathrm{0}\:{is}\:{the}\:\:{only}\:{integer}>> \\ $$$${Where}\:{is}\:{the}\:{error}? \\ $$$$\left.\mathrm{4}\right)\:\:\:{let}\:{consider}\:{n}\:{an}\:{integer}\:{different}\:{of}\:{zero}\:{and}\:\:{f}\left({n}\right)={nln}\left({n}\right) \\ $$$${we}\:{have}\:\:\frac{{df}}{{dn}}={ln}\left({n}\right)+\mathrm{1}\:\:\:\:\left(\bullet\right) \\ $$$${Likewise}\:{f}\left({n}\right)={ln}\left({n}^{{n}} \right)\:\:\:{and}\:{we}\:{know}\:{that} \\ $$$${n}^{{n}} ={n}×{n}×{n}×……×{n}\:\:\left({n}\:{times}\right) \\ $$$${So}\:\:{f}\left({n}\right)={ln}\left({n}\right)+{ln}\left({n}\right)+……+{ln}\left({n}\right)\:\:\:\:\:\:\left({n}\:{times}\right) \\ $$$${Now}\:{we}\:{have}\:\:\frac{{df}}{{dn}}=\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{{n}}+….+\frac{\mathrm{1}}{{n}}\:\:\:\:\:\:\left({n}\:\:{times}\right) \\ $$$${So}\:\:\:\frac{{df}}{\:{dn}}=\mathrm{1}\:\:\:\:\:\left(\bullet\bullet\right) \\ $$$${Relation}\:\left(\bullet\right)\:\:{and}\:\:\:\:\left(\bullet\bullet\right)\:\:{give}\:\:\:\: \\ $$$${ln}\left({n}\right)+\mathrm{1}=\mathrm{1}\:\:\:{then}\:\:{ln}\left({n}\right)=\mathrm{0}\:\Rightarrow\:{n}=\mathrm{1} \\ $$$$<<\:{The}\:{logarithm}\:{of}\:\:{all}\:{n}\geqslant\mathrm{1}\:{is}\:{null}\::\:{There}\:{is}\:{no}\:{integer}\:{big}\:{than}\:\mathrm{1}\:>> \\ $$$${Where}\:{is}\:{the}\:{error}? \\ $$$$\left.\mathrm{5}\right)\:{let}\:{consider}\:\:{x}=\mathrm{0},\mathrm{999999999}……. \\ $$$${we}\:{ascertain}\:{that}\: \\ $$$$\mathrm{10}{x}=\mathrm{9},\mathrm{999999999}…… \\ $$$${then}\:\:\mathrm{10}{x}=\mathrm{9}+\mathrm{0},\mathrm{999999999}…. \\ $$$${So}\:\:\mathrm{10}{x}=\mathrm{9}+{x}\:\: \\ $$$${finally}\:\:{x}=\mathrm{1} \\ $$$$<<\:\mathrm{0}.\mathrm{9999999999999999}\:…..\:\:{is}\:\:{and}\:{integer}\:>> \\ $$$${Is}\:{there}\:{any}\:{error}? \\ $$$$\left.\mathrm{6}\right){let}\:{consider}\:\:{a}=\frac{\mathrm{26666666666666666}}{\mathrm{66666666666666665}}\: \\ $$$${b}=\frac{\mathrm{999999999999999999999995}}{\mathrm{199999999999999999999999}}\: \\ $$$${In}\:{the}\:{way}\:{to}\:{cancel}\:,\:{if}\:{i}\:{just}\:{remove}\:{one}\:{common} \\ $$$${figure}\:{to}\:{the}\:{numerator}\:{and}\:{to}\:{the}\:{denominator} \\ $$$${And}\:{i}\:{find}\:{a}=\frac{\mathrm{2}}{\mathrm{5}}\:\:{and}\:\:{b}=\frac{\mathrm{5}}{\mathrm{1}}\: \\ $$$${Will}\:{it}\:{be}\:{wrong}?\:{if}\:\:{no}\:,{explain}! \\ $$$$\: \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\: \\ $$