Is the result upper bounded by $E_i(\ln(2)T)$ ?
Edit: where $$E_i(y) = \int_{-\infty}^y \frac{\exp(z)}{z} \mathbb{dz}$$
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What's $E_i$ in your expression? – 5xum Apr 29 '16 at 12:23
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Would a summation by parts work? – jdods Apr 29 '16 at 12:26
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@jdods I didn't get. Please reply with your expression. Thanks! – aroyc Apr 29 '16 at 12:30
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Look up summation by parts or Newton summation. There are formula for finite sums where the summand is the product of sequences which have nice finite sum formula. – jdods Apr 29 '16 at 12:38
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There is a "closed form" involving the incomplete Beta function $- B(2; n+1,0) - i\pi$ – KoA Apr 29 '16 at 12:39
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What is $n$ ? Also the sum must be in $\mathbb{R}$ right ? – aroyc Apr 30 '16 at 17:02
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Obviously the main contribute of $\sum_{t=1}^{T}\frac{2^t}{t}$ comes from the last terms, and
$$\sum_{t=1}^{T}\frac{2^t}{t} = \frac{2^T}{T}\sum_{t=0}^{T-1}\frac{1}{2^{t}(1-\frac{t}{T})}\geq\frac{2^T}{T}\sum_{t=0}^{T-1}\frac{1+\frac{t}{T}}{2^t}=\frac{2^{T+1}}{T}\left(1+\frac{1}{T}+O(2^{-T})\right). $$ To approximate the sum with an integral is another good way to go.
Jack D'Aurizio
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