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P-n-e-1-1-1-2-1-3-1-4-1-n-1-1-n-e-1-1-2-1-3-1-4-1-n-1-1-n-e-k-1-n-1-k-1-k-P-l

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Question Number 175770 by mnjuly1970 last updated on 06/Sep/22
P_n = e^( ((1/1) −(1/2)) +((1/3) −(1/4)) +...+((1/(n−1)) −(1/n))) = e^( (1−(1/2) +(1/3) −(1/4) +...+(1/(n−1)) −(1/n))) = e^( Σ_(k=1) ^n (( (−1 )^( k+1) )/k)) ∴ P = lim_( n→∞) (e^( Σ_(k=1) ^n (((−1)^( k+1) )/k)) ) = e^( lim_( n→∞) ( Σ_(k=1) ^n (((−1)^(k+1) )/k))) = e^( ln(2)) = 2
$$ \\ $$$$\:\:\:\mathrm{P}_{{n}} \:=\:{e}^{\:\left(\frac{\mathrm{1}}{\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{2}}\right)\:+\left(\frac{\mathrm{1}}{\mathrm{3}}\:−\frac{\mathrm{1}}{\mathrm{4}}\right)\:+…+\left(\frac{\mathrm{1}}{{n}−\mathrm{1}}\:−\frac{\mathrm{1}}{{n}}\right)} \\ $$$$\:\:\:\:\:\:\:\:=\:{e}^{\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{3}}\:−\frac{\mathrm{1}}{\mathrm{4}}\:\:+…+\frac{\mathrm{1}}{{n}−\mathrm{1}}\:−\frac{\mathrm{1}}{{n}}\right)} \\ $$$$\:\:\:\:\:\:\:\:=\:{e}^{\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\:\left(−\mathrm{1}\:\right)^{\:{k}+\mathrm{1}} }{{k}}} \\ $$$$\:\:\:\:\:\:\:\:\:\therefore\:\:\mathrm{P}\:=\:{lim}_{\:{n}\rightarrow\infty} \left({e}^{\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\left(−\mathrm{1}\right)^{\:{k}+\mathrm{1}} }{{k}}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{e}^{\:{lim}_{\:{n}\rightarrow\infty} \left(\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} }{{k}}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{e}^{\:{ln}\left(\mathrm{2}\right)} =\:\mathrm{2} \\ $$

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