Let c =∑n=1^∞ 1/n(2^n − 1) = 1 +1/6+1/21+1/60+ · · ·. Show that e^c =2/1·4/3·8/7·16/15· · · · .

Question

Let
[latex]c =\sum\limits_{n=1}^{\infty} \frac{1}{n(2^n − 1) }= 1 +\frac{1}{6}+\frac{1}{21}+\frac{1}{60}+ · · ·.[/latex]
Show that
[latex]e^c =\frac{2}{1}·\frac{4}{3}·\frac{8}{7}·\frac{16}{15}· · · · .[/latex]

Accepted Answer

Let [latex]P_k[/latex] be the partial product

[latex]{\frac{2}{1}}\cdot{\frac{4}{3}}\cdot{\frac{8}{7}}\cdot\cdot\cdot{\frac{2^{k}}{2^{k}-1}}.[/latex]

We want to show that

[latex]\lim_{k o\infty}P_{k}=e^{c},[/latex]

where

[latex]c=\sum\limits_{n=1}^{\infty}{\frac{1}{n(2^{n}-1)}}\,.[/latex]

By continuity of the exponential and logarithm functions, this is equivalent to showing that [latex]\lim_{k→\infty} \ln(P_k) = c[/latex]. Now

[latex]\ln(P_{k})=\ln\left({\frac{2}{1}}\cdot{\frac{4}{3}}\cdot{\frac{8}{7}}\cdot\cdot\cdot{\frac{2^{k}}{2^{k}-1}} ight)\=\ln\left({\frac{2}{1}} ight)+\ln\left({\frac{4}{3}} ight)+\ln\left({\frac{8}{7}} ight)+\cdot\cdot\cdot+\ln\left({\frac{2^{k}}{2^{k}-1}} ight)\=-\ln\left({\frac{1}{2}} ight)-\ln\left({\frac{3}{4}} ight)-\ln\left({\frac{7}{8}} ight)-\cdot\cdot\cdot-\ln\left({\frac{2^{k}-1}{2^{k}}} ight)\=-\ln\left(1-{\frac{1}{2}} ight)-\ln\left(1-{\frac{1}{4}} ight)-\ln\left(1-{\frac{1}{8}} ight)-\cdot\cdot\cdot-\ln\left(1-{\frac{1}{2^{8}}} ight)\=\sum\limits_{i=1}^{k}-\ln\left(1-{\frac{1}{2^{i}}} ight).[/latex]

Using the Maclaurin expansion (Taylor expansion about 0)

[latex]-\ln(1-x)=x+{\frac{x^{2}}{2}}+{\frac{x^{3}}{3}}+\cdot\cdot\cdot=\sum\limits_{n=1}^{\infty}{\frac{x^{n}}{n}}\qquad(|x|\lt 1),[/latex]

we get

[latex]\ln(P_{k})=\sum\limits_{i=1}^{k}\sum\limits_{n=1}^{\infty}{\frac{(1/2^{i})^{n}}{n}}=\sum_{n=1}^{\infty}\sum\limits_{i=1}^{k}{\frac{1}{n2^{i n}}}=\sum\limits_{n=1}^{\infty}{\frac{1}{n}}\,\sum\limits_{i=1}^{k}{\frac{1}{(2^{n})^{i}}}\,.[/latex]

Then, because

[latex]\sum\limits_{i=1}^{k}{\frac{1}{(2^{n})^{i}}}={\frac{1}{2^{n}}}+{\frac{1}{(2^{n})^{2}}}+\cdot\cdot\cdot+{\frac{1}{(2^{n})^{k}}}[/latex]

is a finite geometric series with sum

[latex]{\frac{1}{2^{n}}}\cdot{\frac{1-(1/2^{n})^{k}}{1-1/2^{n}}}={\frac{1-(1/2^{n})^{k}}{2^{n}-1}}\,,[/latex]

we have

[latex]\ln(P_{k})=\sum\limits_{n=1}^{\infty}{\frac{1-(1/2^{n})^{k}}{n(2^{n}-1)}}=\sum\limits_{n=1}^{\infty}{\frac{1}{n(2^{n}-1)}}-\sum\limits_{n=1}^{\infty}{\frac{1}{n(2^{n}-1)2^{n k}}}[/latex]

since the latter two series converge. From the estimate

[latex]\sum\limits_{n=1}^{\infty}{\frac{1}{n(2^{n}-1)2^{n k}}}\lt {\frac{1}{2^{k}}}\sum\limits_{n=1}^{\infty}{\frac{1}{n(2^{n}-1)}}[/latex]

we conclude that [latex]\lim_{k o\infty}\ln{(P_{k})}=\sum\limits_{n=1}^{\infty}{\frac{1}{n(2^{n}-1)}}=c,[/latex] as claimed.

Question P.177: Let c =∑n=1^∞ 1/n(2^n − 1) = 1 +1/6+1/21+1/60+ · · ·. Show t......

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