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It is not difficult to show that $$ S_m = \sum_{d=1}^m \frac {2^d}{\sqrt{d}} = \frac {2^{m+1}}{\sqrt{m}} (1+O(\log(m)/m)). $$

But I was wondering if it is possible to do such things using computer? I tried maple's aysmpt function using the following code:

f := m -> sum(2^d/sqrt(d), d = 1 .. m);
asympt(f(x), x);

It does not work. Does there exist programs that can do such jobs?

The correct relative error term is $O(1/m)$. I didn't get this using computer, but numerical computation does hint the correct order.

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    Mathematica fails too: unless you consider `-2^(1 + n) LerchPhi[2, 1/2, 1 + n] + PolyLog[1/2, 2]' a step in the right direction. – 2017-01-25
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    @ClementC. Can't Mathematica expand this some properly? – 2017-01-25
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    @tired I tried, but didn't succeed. I am no Mathematica wizard, though. – 2017-01-25

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