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Question Number 6535 by Temp last updated on 01/Jul/16
Σ_(x=1) ^∞ (1/x) = 1+(1/2)+(1/3)+(1/4)+(1/5)+... Is the following true? ∴Σ(1/x)=Σ((1/(2x))+(1/(2x−1))) =Σ(((2x−1+2x)/(2x(2x−1)))) =Σ(((2x−2+1+2x)/(2x(2x−1)))) =Σ(((4x−2)/(2x(2x−1)))+(1/(2x(2x−1)))) =Σ(((2(2x−1))/(2x(2x−1)))+(1/(2x(2x−1)))) =Σ((1/x)+(1/(2x(2x−1)))) ∴ Σ_(x=1) ^∞ (1/x) =Σ_(x=1) ^∞ (1/x) +Σ_(x=1) ^∞ (1/(2x(2x−1)))
$$\underset{{x}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{x}}\:=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}}+… \\ $$$$\boldsymbol{\mathrm{Is}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{following}}\:\boldsymbol{\mathrm{true}}? \\ $$$$\therefore\Sigma\frac{\mathrm{1}}{{x}}=\Sigma\left(\frac{\mathrm{1}}{\mathrm{2}{x}}+\frac{\mathrm{1}}{\mathrm{2}{x}−\mathrm{1}}\right) \\ $$$$=\Sigma\left(\frac{\mathrm{2}{x}−\mathrm{1}+\mathrm{2}{x}}{\mathrm{2}{x}\left(\mathrm{2}{x}−\mathrm{1}\right)}\right) \\ $$$$=\Sigma\left(\frac{\mathrm{2}{x}−\mathrm{2}+\mathrm{1}+\mathrm{2}{x}}{\mathrm{2}{x}\left(\mathrm{2}{x}−\mathrm{1}\right)}\right) \\ $$$$=\Sigma\left(\frac{\mathrm{4}{x}−\mathrm{2}}{\mathrm{2}{x}\left(\mathrm{2}{x}−\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2}{x}\left(\mathrm{2}{x}−\mathrm{1}\right)}\right) \\ $$$$=\Sigma\left(\frac{\mathrm{2}\left(\mathrm{2}{x}−\mathrm{1}\right)}{\mathrm{2}{x}\left(\mathrm{2}{x}−\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2}{x}\left(\mathrm{2}{x}−\mathrm{1}\right)}\right) \\ $$$$=\Sigma\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{\mathrm{2}{x}\left(\mathrm{2}{x}−\mathrm{1}\right)}\right) \\ $$$$\therefore\:\underset{{x}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{x}}\:=\underset{{x}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{x}}\:+\underset{{x}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{x}\left(\mathrm{2}{x}−\mathrm{1}\right)} \\ $$
Commented by Temp last updated on 01/Jul/16
Note: Σ_(x=1) ^∞ (1/(2x(2x−1)))=Σ_(x=1) ^∞ ((1/(2x−1))−(1/(2x))) Σ_(x=1) ^∞ ((1/(2x−1))−(1/(2x)))=((1/1)−(1/2))+((1/3)−(1/4))+... Σ_(x=1) ^∞ ((1/(2x−1))−(1/(2x)))=1−(1/2)+(1/3)−(1/4)+(1/5)−(1/6)+... =ln(2) ∴ Σ_(x=1) ^∞ (1/x) = Σ_(x=1) ^∞ (1/x) + ln(2) How can this be true?
$$\boldsymbol{\mathrm{Note}}: \\ $$$$\underset{{x}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{x}\left(\mathrm{2}{x}−\mathrm{1}\right)}=\underset{{x}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}{x}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{x}}\right) \\ $$$$\underset{{x}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}{x}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{x}}\right)=\left(\frac{\mathrm{1}}{\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\left(\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}\right)+… \\ $$$$\underset{{x}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}{x}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{x}}\right)=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{6}}+… \\ $$$$=\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$ \\ $$$$\therefore\:\underset{{x}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{x}}\:=\:\underset{{x}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{x}}\:+\:\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$\mathrm{How}\:\mathrm{can}\:\mathrm{this}\:\mathrm{be}\:\mathrm{true}? \\ $$
Commented by prakash jain last updated on 01/Jul/16
Σ(1/n) is divergent hence splitting& rearrangmnts are not valid.
$$\Sigma\frac{\mathrm{1}}{{n}}\:\mathrm{is}\:\mathrm{divergent}\:\mathrm{hence}\:\mathrm{splitting\&}\:\mathrm{rearrangmnts} \\ $$$$\mathrm{are}\:\mathrm{not}\:\mathrm{valid}. \\ $$
Commented by prakash jain last updated on 01/Jul/16
a_n =b_n +c_n then Σ a_n =Σ b_n +Σ c_n is valid only for absolutely convergent series.
$${a}_{{n}} ={b}_{{n}} +{c}_{{n}} \\ $$$$\mathrm{then} \\ $$$$\Sigma\:{a}_{{n}} =\Sigma\:{b}_{{n}} +\Sigma\:{c}_{{n}} \:\mathrm{is}\:\mathrm{valid}\:\mathrm{only}\:\mathrm{for} \\ $$$$\mathrm{absolutely}\:\mathrm{convergent}\:\mathrm{series}. \\ $$
Commented by Temp last updated on 02/Jul/16
I am not certain this is completely incorrect. If you solve both sides: ∞=∞+ln(2) True Although ∞−∞≠0 ∴ln(2)≠0
$$\mathrm{I}\:\mathrm{am}\:\mathrm{not}\:\mathrm{certain}\:\mathrm{this}\:\mathrm{is}\:\mathrm{completely}\:\mathrm{incorrect}. \\ $$$$\mathrm{If}\:\mathrm{you}\:\mathrm{solve}\:\mathrm{both}\:\mathrm{sides}: \\ $$$$\infty=\infty+\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$\mathrm{True} \\ $$$$ \\ $$$$\mathrm{Although} \\ $$$$\infty−\infty\neq\mathrm{0} \\ $$$$\therefore\mathrm{ln}\left(\mathrm{2}\right)\neq\mathrm{0} \\ $$

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