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I tried to compute the series expansion of this equation

r1 = 1/2 (1 + 2^(-1 - k) (1 + (3/5)^(1 + k))^(1 + k))
Series[r1, {k, Infinity, 1}] // Normal // Expand 

which gives ouput

1/2 + 2^(-2 - k) (1 + (5/3)^(-1 - k))^k +  3/20 (1 + (5/3)^(-1 - k))^k E^(k (-Log[5/3] - Log[2]))

This is not particularly useful. A clearer result would be

1/2 + 1/2^(k+2) + O(k/2^(2 k)),

which can be verified either manually or using Maple. Is there a better way to do this type of series expansion in Mathematica?

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  • $\begingroup$ What version are you using? When I run your code, it basically spits back the original function without doing any expansion. $\endgroup$ – march Dec 18 '18 at 16:48
  • $\begingroup$ I use version Mathematica 11.3 and Maple 18. $\endgroup$ – ablmf Dec 19 '18 at 9:50
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If you transform your equation to

2 r1-1==1/2^(k + 1) (1 + (3/5)^(k + 1))^(k + 1)

and use the binomial expansion for the last term you'll get

2 r1-1~ 1/2^(k + 1)(1+3/5(k+1)+1/2 (3/5)^2 k(k+1)+O[k^3])

That's the expansion you found!

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