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Question Number 3077 by Filup last updated on 04/Dec/15
Read the following: S=Σ_(i=1) ^∞ n^(i!) S=n^1 +n^(1×2) +n^(1×2×3) +... S=n^1 (1+n^2 +n^(2×3) +...) S=n(1+S) ∴S=(n/(1−n)) How can this possibly be correct?!
$$\mathrm{Read}\:\mathrm{the}\:\mathrm{following}: \\ $$$$ \\ $$$${S}=\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}{n}^{{i}!} \\ $$$${S}={n}^{\mathrm{1}} +{n}^{\mathrm{1}×\mathrm{2}} +{n}^{\mathrm{1}×\mathrm{2}×\mathrm{3}} +… \\ $$$${S}={n}^{\mathrm{1}} \left(\mathrm{1}+{n}^{\mathrm{2}} +{n}^{\mathrm{2}×\mathrm{3}} +…\right) \\ $$$${S}={n}\left(\mathrm{1}+{S}\right) \\ $$$$\therefore{S}=\frac{{n}}{\mathrm{1}−{n}} \\ $$$$ \\ $$$$\mathrm{How}\:\mathrm{can}\:\mathrm{this}\:\mathrm{possibly}\:\mathrm{be}\:\mathrm{correct}?! \\ $$
Commented by Filup last updated on 04/Dec/15
For n=1 S=1+1+1+....=(1/0) =undefined For n=2 S=2+4+64+...=(2/(−1)) =−2 For n=3 S=3+9+729+...=(3/(−2)) etc.
$$\mathrm{For}\:{n}=\mathrm{1} \\ $$$${S}=\mathrm{1}+\mathrm{1}+\mathrm{1}+….=\frac{\mathrm{1}}{\mathrm{0}} \\ $$$$=\mathrm{undefined} \\ $$$$ \\ $$$$\mathrm{For}\:{n}=\mathrm{2} \\ $$$${S}=\mathrm{2}+\mathrm{4}+\mathrm{64}+…=\frac{\mathrm{2}}{−\mathrm{1}} \\ $$$$=−\mathrm{2} \\ $$$$ \\ $$$$\mathrm{For}\:{n}=\mathrm{3} \\ $$$${S}=\mathrm{3}+\mathrm{9}+\mathrm{729}+…=\frac{\mathrm{3}}{−\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{etc}. \\ $$
Commented by Filup last updated on 04/Dec/15
The above comment seems incorrect. If 1+1+1+... is undefined, 2+4+64+... should also be such because it grows exponentially faster than the prior example
$$\mathrm{The}\:\mathrm{above}\:\mathrm{comment}\:\mathrm{seems}\:\mathrm{incorrect}. \\ $$$$\mathrm{If}\:\:\mathrm{1}+\mathrm{1}+\mathrm{1}+…\:\:\mathrm{is}\:\mathrm{undefined}, \\ $$$$\mathrm{2}+\mathrm{4}+\mathrm{64}+…\:\:\mathrm{should}\:\mathrm{also}\:\mathrm{be}\:\mathrm{such}\:\mathrm{because} \\ $$$$\mathrm{it}\:\mathrm{grows}\:\mathrm{exponentially}\:\mathrm{faster}\:\mathrm{than} \\ $$$$\mathrm{the}\:\mathrm{prior}\:\mathrm{example} \\ $$
Commented by Rasheed Soomro last updated on 04/Dec/15
Your 4th line seems incorrect.See my answer.
$${Your}\:\mathrm{4}{th}\:{line}\:{seems}\:{incorrect}.{See}\:{my}\:{answer}. \\ $$
Commented by prakash jain last updated on 04/Dec/15
The argument in question only apply to geometric series S=Σ_(k=0) ^∞ a∙r^k S=a+rS S=(a/(1−r)) ∣r∣<1
$$\mathrm{The}\:\mathrm{argument}\:\mathrm{in}\:\mathrm{question}\:\mathrm{only}\:\mathrm{apply}\:\mathrm{to} \\ $$$$\mathrm{geometric}\:\mathrm{series} \\ $$$${S}=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{a}\centerdot{r}^{{k}} \\ $$$${S}={a}+{rS} \\ $$$${S}=\frac{{a}}{\mathrm{1}−{r}} \\ $$$$\mid{r}\mid<\mathrm{1} \\ $$
Answered by Rasheed Soomro last updated on 04/Dec/15
S=n^1 +n^(1×2) +n^(1×2×3) +... S=n^1 (1+n^2 +n^(2×3) +...)??? S=n^1 (1+n^(1×2−1) +n^(1×2×3−1) +n^(1×2×3×4−1) ...) S=n^1 (1+n^1 +n^5 +n^(23) ...)
$${S}={n}^{\mathrm{1}} +{n}^{\mathrm{1}×\mathrm{2}} +{n}^{\mathrm{1}×\mathrm{2}×\mathrm{3}} +… \\ $$$${S}={n}^{\mathrm{1}} \left(\mathrm{1}+{n}^{\mathrm{2}} +{n}^{\mathrm{2}×\mathrm{3}} +…\right)??? \\ $$$${S}={n}^{\mathrm{1}} \left(\mathrm{1}+{n}^{\mathrm{1}×\mathrm{2}−\mathrm{1}} +{n}^{\mathrm{1}×\mathrm{2}×\mathrm{3}−\mathrm{1}} +{n}^{\mathrm{1}×\mathrm{2}×\mathrm{3}×\mathrm{4}−\mathrm{1}} …\right) \\ $$$${S}={n}^{\mathrm{1}} \left(\mathrm{1}+{n}^{\mathrm{1}} +{n}^{\mathrm{5}} +{n}^{\mathrm{23}} …\right) \\ $$$$ \\ $$
Commented by Filup last updated on 04/Dec/15
Ah! Youre right! n^(1×2×3) =(n^(2×3) )^1 =(n^2 )^3 not n^(1×2×3) =n^2 (n^3 )
$$\mathrm{Ah}!\:\mathrm{Youre}\:\mathrm{right}! \\ $$$${n}^{\mathrm{1}×\mathrm{2}×\mathrm{3}} =\left({n}^{\mathrm{2}×\mathrm{3}} \right)^{\mathrm{1}} =\left({n}^{\mathrm{2}} \right)^{\mathrm{3}} \\ $$$$\mathrm{not}\:{n}^{\mathrm{1}×\mathrm{2}×\mathrm{3}} ={n}^{\mathrm{2}} \left({n}^{\mathrm{3}} \right) \\ $$

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