D(g) Summation

Calculus Level 5

D n ( g ) = ∑ k = 0 n 2 k g 2 k g ( 1 + g 2 k ) \displaystyle{{ D }_{ n }\left( g \right) =\sum _{ k=0 }^{ n }{ \cfrac { { 2 }^{ k }{ g }^{ { 2 }^{ k } } }{ g(1+{ g }^{ { 2 }^{ k } }) } } }

Define above summation such that ∣ g ∣ < 1 \displaystyle{\left| g \right| <1} .

lim ⁡ n → ∞ D n ( 1 2015 ) = a b \displaystyle{\lim _{ n\rightarrow \infty }{ { D }_{ n }\left( \cfrac { 1 }{ 2015 } \right) } =\cfrac { a }{ b } }

Find 2 a − b 2a-b

Details and assumptions

∙ \bullet Here a , b a,b are co-primes.

Original


The answer is 2016.

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Deepanshu Gupta by Deepanshu Gupta , 25, India

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1 solution

Kartik Sharma
Mar 14, 2015

D n ( g ) = ∑ k = 0 n 2 k ( g ) 2 k ( 1 + ( g ) 2 k ) g \displaystyle {D}_{n}(g) = \sum_{k=0}^{n}{\frac{{2}^{k}{(g)}^{{2}^{k}}}{(1+{(g)}^{{2}^{k}})g}}

D n ( 1 x ) = x 1 g ∑ k = 0 n 2 k x 2 k + 1 \displaystyle {D}_{n}(\frac{1}{x}) = x\frac{1}{g}\sum_{k=0}^{n}{\frac{{2}^{k}}{{x}^{{2}^{k}}+1}}

D n ( 1 x ) = x ∑ k = 0 n 2 k ( x 2 k − 1 ) x 2 k + 1 − 1 \displaystyle {D}_{n}(\frac{1}{x}) = x\sum_{k=0}^{n}{\frac{{2}^{k}({x}^{{2}^{k}} - 1)}{{x}^{{2}^{k+1}} -1}}

D n ( 1 x ) = x ∑ k = 0 n 2 k x 2 k x 2 k + 1 − 1 − 2 k x 2 k + 1 − 1 \displaystyle {D}_{n}(\frac{1}{x}) = x\sum_{k=0}^{n}{\frac{{2}^{k}{x}^{{2}^{k}}}{{x}^{{2}^{k+1}} -1} - \frac{{2}^{k}}{{x}^{{2}^{k+1}} -1}}

Partial fraction for this sum - x 2 k x 2 k + 1 − 1 = 1 x 2 k − 1 − 1 x 2 k + 1 − 1 \displaystyle \frac{{x}^{{2}^{k}}}{{x}^{{2}^{k+1}} -1} = \frac{1}{{x}^{{2}^{k}}-1} - \frac{1}{{x}^{{2}^{k+1}}-1}

Hence,

D n ( 1 x ) = x ∑ k = 0 n 2 k x 2 k − 1 − 2 k x 2 k + 1 − 1 − 2 k x 2 k + 1 − 1 \displaystyle {D}_{n}(\frac{1}{x}) = x\sum_{k=0}^{n}{\frac{{2}^{k}}{{x}^{{2}^{k}}-1} - \frac{{2}^{k}}{{x}^{{2}^{k+1}}-1} - \frac{{2}^{k}}{{x}^{{2}^{k+1}} -1}}

Therefore,

D n ( 1 x ) = x ∑ k = 0 n 2 k x 2 k − 1 − 2 k + 1 x 2 k + 1 − 1 \displaystyle {D}_{n}(\frac{1}{x}) = x\sum_{k=0}^{n}{\frac{{2}^{k}}{{x}^{{2}^{k}}-1} - \frac{{2}^{k+1}}{{x}^{{2}^{k+1}}-1}}

which telescopes to x x − 1 − x 2 n + 1 x 2 n + 1 − 1 \displaystyle \frac{x}{x-1} - \frac{x{2}^{n+1}}{{x}^{{2}^{n+1}}-1}

l i m n − > ∞ D n ( 1 2015 ) = 2015 2014 \displaystyle {lim}_{n->\infty} {D}_{n}(\frac{1}{2015}) = \frac{2015}{2014}

as the 2nd term will come to 0.

Hence the answer becomes 2015 2014 \displaystyle \boxed{\frac{2015}{2014}}

16 Helpful 1 Interesting 1 Brilliant 0 Confused

wow .. Does someone notice that this question includes 3 years , 2014 ,2015 and 2016 .. It is interesting answer.

Nishu sharma - 6 years, 1 month ago

The main thing about the problem was it looked hard while it is easy. I too at first overthought it. @Deepanshu Gupta Nice problem! Thanks!

Kartik Sharma - 6 years, 3 months ago

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Nice Solution. Actually I have created this question by keeping in mind that question includes Integration . I have Posted Solution also previously But while editing I have accidently deleted almost all latex code. So I deleted that completly. My Intention was : D ( x ) = ∑ n x n − 1 1 + x n ∫ D ( x ) = ∑ ln ⁡ ( 1 + x n ) D(x)=\sum { \cfrac { n{ x }^{ n-1 } }{ 1+{ x }^{ n } } } \\ \int { D(x) } =\sum { \ln { \left( 1+{ x }^{ n } \right) } } something like that. But Really Your approach is Nice. +1 .

Deepanshu Gupta - 6 years, 3 months ago

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First thought which came in my mind was the same as yours but then I just stuck.

Kartik Sharma - 6 years, 3 months ago

Hi , if you delete your own solution , are you able to repost another solution ? I am not able to do so .

P.S. I am asking you since you are a mod !

Thanks :)

A Former Brilliant Member - 6 years, 3 months ago

Did it by integration. @Deepanshu Gupta

Sudeep Salgia - 6 years, 2 months ago

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@Sudeep Salgia – then do post it plz , if you are not busy ! So that others can learn from that. @Sudeep Salgia

Deepanshu Gupta - 6 years, 2 months ago

i'll be glad if you edit your solution as the 2nd step you have typed wrong. that 1/g is extra even though it is self understood , plz edit it.

A Former Brilliant Member - 4 years, 9 months ago

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