Remembering Ramanujan

Calculus Level 5

$\mathfrak{G}=1-\displaystyle \sum^{\infty}_{n=1} \dfrac{1}{1729^n n(n+1)}$

If the value of $\mathfrak{G}$ is in the form $\ln \left(1+\frac{1}{n}\right)^n$ , find $n$ .

The answer is 1728.

1 solution

Guilherme Niedu
Feb 25, 2017

$\large \displaystyle = 1 - \sum_{n=0}^{\infty} \Big (\frac{1}{1729} \Big )^n \cdot \Big (\frac1n - \frac{1}{n+1}\Big )$

$\large \displaystyle = 1 - \Big (\frac{1}{1729} \Big ) + \frac12 \cdot \Big[\Big (\frac{1}{1729} \Big ) - \Big (\frac{1}{1729} \Big )^2 \Big] + \frac13 \cdot \Big[\Big (\frac{1}{1729} \Big )^2 - \Big (\frac{1}{1729} \Big )^3 \Big] + ...$

$\large \displaystyle = \Big( 1 - \frac{1}{1729} \Big )\cdot \Big (1 + \frac{1/1729}{2} + \frac{(1/1729)^2}{3} +... \Big )$

$\large \displaystyle = ( 1728)\cdot \Big (1/1729 + \frac{(1/1729)^2}{2} + \frac{(1/1729)^3}{3} +... \Big )$

$\large \displaystyle = ( 1728)\cdot - \ln \Big (1 - \frac{1}{1729}\Big)$

$\large \displaystyle = ( 1728)\cdot \ln \Big [ \Big (\frac{1728}{1729} \Big) ^ {-1} \Big]$

$\large \displaystyle = ( 1728)\cdot \ln \Big (\frac{1729}{1728} \Big)$

$\large \displaystyle = ( 1728)\cdot \ln \Big (1 + \frac{1}{1728} \Big)$

$\large \displaystyle = \ln \Big [ \Big (1 + \frac{1}{1728} \Big)^{1728} \Big]$

So, $\large \displaystyle \color{#3D99F6} \boxed{n = 1728}$

Cleverly executed by noticing the series expansion of $\ln(1-x)$ .

P.S. Remember to put a backslash before the natural log operator (\ln), that would make it look like an operator.

Tapas Mazumdar - 4 years, 3 months ago

Remembering Ramanujan

The answer is 1728.

1 solution

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