Question Number 177099 by mokys last updated on 30/Sep/22
Answered by TheHoneyCat last updated on 04/Oct/22
![taking the log of your product L=Σ_(n=0) ^∞ x^n (ln(1+x^(n+1) )−ln(1+x^n )) =(1/x)Σ_(n=0) ^∞ x^(n+1) ln(1+x^(n+1) )−Σ_(n=0) ^∞ ln(1+x^n ) =(1/x)(Σ_(n=0) ^∞ x^n ln(1+x^n )−ln2)−Σ_(n=0) ^∞ ln(1+x^n ) =((−ln2)/x)+((1/x)−1)Σ_(n=0) ^∞ x^n ln(1+x^n ) so now we just need to take care of that series. 0<ln(1+x^n )<ln2 so 0>((1/x)−1)Σ…>((1−x)/x) (1/(x−1))ln2=((−ln2)/x) so this limit is indeed wel defined. so going back to the exponential the limit is in [1/4, 1/2] calling l the excat value of the series, your result is 1/2^l I gave up getting the exact result. But it′s a classic one. You can find it in calculus books. I fear it has no explicit value thought. If my memory is correct the result is 1/2^α where alpha is a famous constant that appears in the asympotic study of log but my meroy is rarely correct. ;−)](https://www.tinkutara.com/question/Q177425.png)
$$\mathrm{taking}\:\mathrm{the}\:\mathrm{log}\:\mathrm{of}\:\mathrm{your}\:\mathrm{product} \\ $$$$ \\ $$$$\mathrm{L}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{{n}} \left(\mathrm{ln}\left(\mathrm{1}+{x}^{{n}+\mathrm{1}} \right)−\mathrm{ln}\left(\mathrm{1}+{x}^{{n}} \right)\right) \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{{x}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{{n}+\mathrm{1}} \mathrm{ln}\left(\mathrm{1}+{x}^{{n}+\mathrm{1}} \right)−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{ln}\left(\mathrm{1}+{x}^{{n}} \right) \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{{x}}\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{{n}} \mathrm{ln}\left(\mathrm{1}+{x}^{{n}} \right)−\mathrm{ln2}\right)−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{ln}\left(\mathrm{1}+{x}^{{n}} \right) \\ $$$$\:\:\:\:=\frac{−\mathrm{ln2}}{{x}}+\left(\frac{\mathrm{1}}{{x}}−\mathrm{1}\right)\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{{n}} \mathrm{ln}\left(\mathrm{1}+{x}^{{n}} \right) \\ $$$$\mathrm{so}\:\mathrm{now}\:\mathrm{we}\:\mathrm{just}\:\mathrm{need}\:\mathrm{to}\:\mathrm{take}\:\mathrm{care}\:\mathrm{of}\:\mathrm{that}\:\mathrm{series}. \\ $$$$\mathrm{0}<\mathrm{ln}\left(\mathrm{1}+{x}^{{n}} \right)<\mathrm{ln2} \\ $$$$\mathrm{so} \\ $$$$\mathrm{0}>\left(\frac{\mathrm{1}}{{x}}−\mathrm{1}\right)\Sigma\ldots>\frac{\mathrm{1}−{x}}{{x}}\:\frac{\mathrm{1}}{{x}−\mathrm{1}}\mathrm{ln2}=\frac{−\mathrm{ln2}}{{x}} \\ $$$$\mathrm{so}\:\mathrm{this}\:\mathrm{limit}\:\mathrm{is}\:\mathrm{indeed}\:\mathrm{wel}\:\mathrm{defined}. \\ $$$$\mathrm{so}\:\mathrm{going}\:\mathrm{back}\:\mathrm{to}\:\mathrm{the}\:\mathrm{exponential}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{is}\:\mathrm{in} \\ $$$$\left[\mathrm{1}/\mathrm{4},\:\mathrm{1}/\mathrm{2}\right] \\ $$$$\mathrm{calling}\:{l}\:\mathrm{the}\:\mathrm{excat}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{series},\:\mathrm{your} \\ $$$$\mathrm{result}\:\mathrm{is}\:\mathrm{1}/\mathrm{2}^{{l}} \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{gave}\:\mathrm{up}\:\mathrm{getting}\:\mathrm{the}\:\mathrm{exact}\:\mathrm{result}. \\ $$$$\mathrm{But}\:\mathrm{it}'\mathrm{s}\:\mathrm{a}\:\mathrm{classic}\:\mathrm{one}.\:\mathrm{You}\:\mathrm{can}\:\mathrm{find}\:\mathrm{it}\:\mathrm{in} \\ $$$$\mathrm{calculus}\:\mathrm{books}. \\ $$$$\mathrm{I}\:\mathrm{fear}\:\mathrm{it}\:\mathrm{has}\:\mathrm{no}\:\mathrm{explicit}\:\mathrm{value}\:\mathrm{thought}. \\ $$$$\mathrm{If}\:\mathrm{my}\:\mathrm{memory}\:\mathrm{is}\:\mathrm{correct}\:\mathrm{the}\:\mathrm{result}\:\mathrm{is} \\ $$$$\mathrm{1}/\mathrm{2}^{\alpha} \:\mathrm{where}\:\mathrm{alpha}\:\mathrm{is}\:\mathrm{a}\:\mathrm{famous}\:\mathrm{constant} \\ $$$$\mathrm{that}\:\mathrm{appears}\:\mathrm{in}\:\mathrm{the}\:\mathrm{asympotic}\:\mathrm{study}\:\mathrm{of}\:\mathrm{log} \\ $$$$\mathrm{but}\:\mathrm{my}\:\mathrm{meroy}\:\mathrm{is}\:\mathrm{rarely}\:\mathrm{correct}. \\ $$$$\left.;−\right) \\ $$