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!


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Question Number 67651 by ~ À ® @ 237 ~ last updated on 29/Aug/19

$1) L e t c o n s i d e r S = \underset{n = 0}{\sum^{\infty}} n a n d T = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n + 1} n$ $W e k n o w t h a t \forall x \in] - 1; 1]$ $\underset{n = 0}{\sum^{\infty}} {(- x)}^{n} = \frac{1}{1 + x},$ $t h e n a f t e r d e r i v a t i n g$ $\frac{1}{{(1 + x)}^{2}} = \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n + 1} {n x}^{n - 1}$ $f o r x = 1, w e g e t T = \frac{1}{4}$ $N o w l e t a s c e r t a i n s o m e t h i n g$ $T = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{2 n + 2} (2 n + 1) + \underset{n = 0}{\sum^{\infty}} {(- 1)}^{2 n + 1} (2 n)$ $= \underset{n = 0}{\sum^{\infty}} (2 n + 1) - 2 S (∙)$ $k n o w i n g t h a t S = \underset{n = 0}{\sum^{\infty}} (2 n) + \underset{n = 0}{\sum^{\infty}} (2 n + 1)$ $S o \underset{n = 0}{\sum^{\infty}} (2 n + 1) = S - 2 S$ $W h e n r e p l a c i n g t h a t v a l u e i n (∙)$ $w e g e t$ $T = (S - 2 S) - 2 S$ $I f i c o n c l u d e t h a t T = - 3 S$ $a n d f i n a l l y f i n d S = - \frac{T}{3} = - \frac{1}{12}$ $w h e r e w i l l t h e m i s t a k e b e ?$ $2) L e t c o n s i d e r K = 1 + \frac{2020}{2019} + {(\frac{2020}{2019})}^{2} + {(\frac{2020}{2019})}^{3} + . . . . . . + . . .$ $\frac{2020}{2019} K = \frac{2020}{2019} + {(\frac{2020}{2019})}^{2} + {(\frac{2020}{2019})}^{3} + . . . .$ $K - 1 = \frac{2020}{2019} + {(\frac{2020}{2019})}^{2} + {(\frac{2020}{2019})}^{3} + . . .$ $S o \frac{2020}{2019} K = K - 1$ $T h e n K = - 2019$ $W h e r e i s t h e e r r o r ?$ $3) L e t c o n s i d e r n a n i n t e g e r$ $W e h a v e$ $0 = n - n$ $= n + (- n) = n + {(- n)}^{2 \times \frac{1}{2}}$ $= n + {[{(- n)}^{2}]}^{\frac{1}{2}} = n + \sqrt{n^{2}} = n + n = 2 n$ $S o << a l l i n t e g e r a r e n u l l : 0 i s t h e o n l y i n t e g e r >>$ $W h e r e i s t h e e r r o r ?$ $4) l e t c o n s i d e r n a n i n t e g e r d i f f e r e n t o f z e r o a n d f (n) = n l n (n)$ $w e h a v e \frac{d f}{d n} = l n (n) + 1 (∙)$ $L i k e w i s e f (n) = l n (n^{n}) a n d w e k n o w t h a t$ $n^{n} = n \times n \times n \times . . . . . . \times n (n t i m e s)$ $S o f (n) = l n (n) + l n (n) + . . . . . . + l n (n) (n t i m e s)$ $N o w w e h a v e \frac{d f}{d n} = \frac{1}{n} + \frac{1}{n} + . . . . + \frac{1}{n} (n t i m e s)$ $S o \frac{d f}{d n} = 1 (∙ ∙)$ $R e l a t i o n (∙) a n d (∙ ∙) g i v e$ $l n (n) + 1 = 1 t h e n l n (n) = 0 \Rightarrow n = 1$ $<< T h e l o g a r i t h m o f a l l n ⩾ 1 i s n u l l : T h e r e i s n o i n t e g e r b i g t h a n 1 >>$ $W h e r e i s t h e e r r o r ?$ $5) l e t c o n s i d e r x = 0, 999999999 . . . . . . .$ $w e a s c e r t a i n t h a t$ $10 x = 9, 999999999 . . . . . .$ $t h e n 10 x = 9 + 0, 999999999 . . . .$ $S o 10 x = 9 + x$ $f i n a l l y x = 1$ $<< 0.9999999999999999 . . . . . i s a n d i n t e g e r >>$ $I s t h e r e a n y e r r o r ?$ $6) l e t c o n s i d e r a = \frac{26666666666666666}{66666666666666665}$ $b = \frac{999999999999999999999995}{199999999999999999999999}$ $I n t h e w a y t o c a n c e l, i f i j u s t r e m o v e o n e c o m m o n$ $f i g u r e t o t h e n u m e r a t o r a n d t o t h e d e n o m i n a t o r$ $A n d i f i n d a = \frac{2}{5} a n d b = \frac{5}{1}$ $W i l l i t b e w r o n g ? i f n o, e x p l a i n!$
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