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1-Let-consider-S-n-0-n-and-T-n-0-1-n-1-n-We-know-that-x-1-1-n-0-x-n-1-1-x-then-after-derivating-1-1-x-2-n-1-1-n-1-nx-n-1-f

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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!
1)LetconsiderS=∑∞n=0nandT=∑∞n=0(−1)n+1nWeknowthat∀x∈]−1;1]∑∞n=0(−x)n=11+x,thenafterderivating1(1+x)2=∑∞n=1(−1)n+1nxn−1forx=1,wegetT=14NowletascertainsomethingT=∑∞n=0(−1)2n+2(2n+1)+∑∞n=0(−1)2n+1(2n)=∑∞n=0(2n+1)−2S(∙)knowingthatS=∑∞n=0(2n)+∑∞n=0(2n+1)So∑∞n=0(2n+1)=S−2SWhenreplacingthatvaluein(∙)wegetT=(S−2S)−2SIficoncludethatT=−3SandfinallyfindS=−T3=−112wherewillthemistakebe?2)LetconsiderK=1+20202019+(20202019)2+(20202019)3+……+…20202019K=20202019+(20202019)2+(20202019)3+….K−1=20202019+(20202019)2+(20202019)3+…So20202019K=K−1ThenK=−2019Whereistheerror?3)LetconsidernanintegerWehave0=n−n=n+(−n)=n+(−n)2×12=n+[(−n)2]12=n+n2=n+n=2nSo<<allintegerarenull:0istheonlyinteger>>Whereistheerror?4)letconsidernanintegerdifferentofzeroandf(n)=nln(n)wehavedfdn=ln(n)+1(∙)Likewisef(n)=ln(nn)andweknowthatnn=n×n×n×……×n(ntimes)Sof(n)=ln(n)+ln(n)+……+ln(n)(ntimes)Nowwehavedfdn=1n+1n+….+1n(ntimes)Sodfdn=1(∙∙)Relation(∙)and(∙∙)giveln(n)+1=1thenln(n)=0⇒n=1<<Thelogarithmofalln⩾1isnull:Thereisnointegerbigthan1>>Whereistheerror?5)letconsiderx=0,999999999…….weascertainthat10x=9,999999999……then10x=9+0,999999999….So10x=9+xfinallyx=1<<0.9999999999999999…..isandinteger>>Isthereanyerror?6)letconsidera=2666666666666666666666666666666665b=999999999999999999999995199999999999999999999999Inthewaytocancel,ifijustremoveonecommonfiguretothenumeratorandtothedenominatorAndifinda=25andb=51Willitbewrong?ifno,explain!

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