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Let$${\left( {1 + \frac{1}{x}} \right)^x} = e\left( {1 - \sum\limits_{k = 1}^\infty {\frac{{{d_k}}}{{{{\left( {\frac{{11}}{{12}} + x} \right)}^k}}}} } \right),$$calculate $${\sum\limits_{k = 1}^\infty {\frac{1}{{1 + d_k^2}}} }.$$ I think we can begin with $$\sum\limits_{k = 1}^\infty {\frac{1}{{1 + d_k^2}}} = \sum\limits_{k = 1}^\infty {\sum\limits_{m = 1}^\infty {{{\left( { - d_k^2} \right)}^{m - 1}}} = } \sum\limits_{m = 1}^\infty {\sum\limits_{k = 1}^\infty {{{\left( { - d_k^2} \right)}^{m - 1}}} } ,$$ because we can't obtain $d_k$, so we can only solve it by assigning the $x$.But I failed!

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    This is a weird question. Where did it come from? How did that 11/12 get there? Why $1/(1+d_k^2)$? – 2017-02-02
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    @martycohen I think it is a weird question,too! It's from a net friend. – 2017-02-02
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    $d_1=1/2$ , $d_2=0$ , $d_3=5/288\approx 0.01736$ , $d_4=139/17280\approx 0,00804$ , $d_5=119/23040\approx 0,00516$ , $d_6=5975/1741824\approx 0,00343$ , ... --- It looks like $d_k\to 0$ for $k\to\infty$ and therefore $\sum\limits_{k=1}^\infty \frac{1}{1+d_k^2}$ divergent. --- Do you have any argument that this series is convergent ? Necessary criterion for the convergence: $d_k\to\infty$ for $k\to\infty$ – 2017-02-09

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