Question-150462

Question Number 150462 by Tawa11 last updated on 12/Aug/21

Answered by puissant last updated on 12/Aug/21

I=Σ_(n=0) ^∞ (−1)^n ln(((2+3n)/(1+3n))) posons U_n =(−1)^n ln(((2+3n)/(1+3n))) ⇒ U_n = (−1)^n ln((((2+3n)/(3n))/((1+3n)/(3n)))) = (−1)^n (ln(1+(2/(3n)))−ln(1+(1/(3n)))) =_(n→+∞) ((2(−1)^n )/(3n))+o((1/n))+(((−1)^n )/(3n))+o((1/n)) d′ou U_n ≈_(n→+∞) (((−1)^n )/n) or Σ(((−1)^n )/n) converge, Alors Σ_(n=0) ^∞ (−1)^n ln(((2+3n)/(1+3n))) = −ln2.....

$$\mathrm{I}=\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{ln}\left(\frac{\mathrm{2}+\mathrm{3n}}{\mathrm{1}+\mathrm{3n}}\right) \\ $$$$\mathrm{posons}\:\mathrm{U}_{\mathrm{n}} =\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{ln}\left(\frac{\mathrm{2}+\mathrm{3n}}{\mathrm{1}+\mathrm{3n}}\right) \\ $$$$\Rightarrow\:\mathrm{U}_{\mathrm{n}} =\:\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{ln}\left(\frac{\frac{\mathrm{2}+\mathrm{3n}}{\mathrm{3n}}}{\frac{\mathrm{1}+\mathrm{3n}}{\mathrm{3n}}}\right)\: \\ $$$$=\:\left(−\mathrm{1}\right)^{\mathrm{n}} \left(\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{3n}}\right)−\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3n}}\right)\right) \\ $$$$\underset{\mathrm{n}\rightarrow+\infty} {=}\frac{\mathrm{2}\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{3n}}+{o}\left(\frac{\mathrm{1}}{\mathrm{n}}\right)+\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{3n}}+{o}\left(\frac{\mathrm{1}}{\mathrm{n}}\right) \\ $$$$\mathrm{d}'\mathrm{ou}\:\mathrm{U}_{\mathrm{n}} \underset{\mathrm{n}\rightarrow+\infty} {\approx}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}} \\ $$$$\mathrm{or}\:\Sigma\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}}\:\mathrm{converge},\:\mathrm{Alors} \\ $$$$\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{ln}\left(\frac{\mathrm{2}+\mathrm{3n}}{\mathrm{1}+\mathrm{3n}}\right)\:=\:−\mathrm{ln2}….. \\ $$

Commented by Tawa11 last updated on 12/Aug/21

Thanks. God bless you sir. I appreciate.

$$\mathrm{Thanks}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$

Answered by aleks041103 last updated on 12/Aug/21

Commented by aleks041103 last updated on 12/Aug/21

$${I}\:{hope}\:{the}\:{image}\:{is}\:{clear}\:{enough}. \\ $$$${In}\:{case}\:{it}'{s}\:{not},\:{ask}\:{me} \\ $$

Commented by Tawa11 last updated on 12/Aug/21

Thanks. God bless you sir. I appreciate. The image is not that clear sir. Thanks for your time.

$$\mathrm{Thanks}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}.\:\:\mathrm{The}\:\mathrm{image}\:\mathrm{is}\:\mathrm{not}\:\mathrm{that}\:\mathrm{clear}\:\mathrm{sir}. \\ $$$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{time}. \\ $$

Commented by Tawa11 last updated on 14/Aug/21

Sir, please help me write it well sir. Thanks.

$$\mathrm{Sir},\:\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{write}\:\mathrm{it}\:\mathrm{well}\:\mathrm{sir}.\:\mathrm{Thanks}. \\ $$

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