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I was trying to solve for the limit as $n\rightarrow\infty$: $\displaystyle A_n=\sum_{i=1}^{n} a_i; a_i=\frac{2^{-i}}{i}$ and I landed at the inequality $\frac{N-1}{N}\left(S_{n+1}^{(N)}-a_{1}\right)<A_{N^{n+1}}-a_{1}<S_n^{(N)}$, for $N>2$; where $\displaystyle S_n^{(N)}=\sum_{k=0}^{n}N^k{a_{N^k}}=\sum_{k=0}^{n}2^{-(N^k)}$.

Whereas I have since found that $\displaystyle\lim_{n\to\infty} A_n=log2$ I am still curious about the series $S_n^{(N)}$

The series is convergent for all $N\in\{2,3,4,\ldots\}$ $$\sum_{k=0}^{\infty} 2^{-(N^k)}$$

Is there an expression for it in terms of $N$? If so, what is it and what are the steps to derive it?

Ashwin B
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  • Let $2^{-N} = x$ and consider that $$(1-x)(1 + x + \cdots + x^r) = 1 - x^{(r+1)}.$$ Now consider that $\lim_{r \to \infty} x^{(r+1)} = 0$, for all $x$ such that $|x| < 1.$ – user2661923 Aug 31 '21 at 06:48
  • Note, the series diverges for $N = 1$: $2^{-(1^x)} = 2^{-1}$. – talbi Aug 31 '21 at 12:07
  • Thank you for pointing it out, yes it diverges for N=1. I have edited it now to exclude the N=1 case – Ashwin B Aug 31 '21 at 12:41
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    @user2661923 I understand how that would work for $\sum_{k=0}^{\infty} {(2^{-N})}^k$, but could you explain how you’d adapt it to suit $\sum_{k=0}^{\infty} {2^{-(N^k)}}$ – Ashwin B Aug 31 '21 at 12:50
  • Try using symbolic regression. There's probably not a closed form but if there is that method will likely find it. – Samuel Aug 31 '21 at 13:31
  • You are right. My previous comment was in error. I actually confused $\displaystyle \sum_{k=0}^\infty 2^{-(N^k)}$ with $\displaystyle \sum_{k=0}^\infty \left(2^{-N}\right)^k.$ – user2661923 Aug 31 '21 at 13:38
  • Please add some problem context and research as seen in this question guide. – Тyma Gaidash Aug 31 '21 at 19:34
  • @AshwinB Would you like me to add anything else to the answer or answer something else? – Тyma Gaidash Sep 02 '21 at 02:46
  • @TymaGaidash I have added context to my question, I hope that is good enough – Ashwin B Sep 03 '21 at 04:29
  • @AshwinB [+1] Yes it is good, but does my answer work for you or should I add more? – Тyma Gaidash Sep 03 '21 at 12:39
  • @TymaGaidash: I am only looking for a closed form expression out of curiosity. Your answer works for me in the sense that it gives me a direction to proceed towards. I do not need you to add anything more. I might suggest pursuing those upper and lower bounds I provided on $S_{n}^{(N)}$ if you are curious too. – Ashwin B Sep 03 '21 at 14:12
  • @AshwinB The notation may be confused with the Stirling number Wolfram notation? Maybe I will take a look at the limits. – Тyma Gaidash Sep 03 '21 at 15:47

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Let’s try to find a pattern within the series. One way is to use the Abel Plana formula. I will replace N with n for simplicity. The result has the Logarithmic Integral function similar to the gamma function. Note the real and imaginary parts are separated in the end. The complex argument function is also used here:

$$\mathrm{\sum_{k=0}^\infty 2^{-n^k}= \sum_{k=0}^\infty \left(\frac12\right)^{n^k}= \sum_{k=0}^\infty e^{{-n^k}\ln(2)}=\frac12 2^{-n^0}+\int_0^\infty 2^{-n^x} dx+i\int _0^\infty\frac{2^{-n^{ix}}-2^{-n^{-ix}}}{e^{2\pi x}-1}dx=\frac14-\frac{li\left(\frac12\right)}{ln(n)}+i\int _0^\infty \frac{1}{e^{2\pi x}-1}\left(2^{-e^{-x arg(n)} cos( x log(n))} cos(log(2) e^{-x arg(n)} sin(x log(n))) - 2^{-e^{x arg(n)} cos(x log(n))} cos(log(2) e^{x arg(n)} sin(x log(n)))+ i (sin(log(2) e^{-x arg(n)} sin( x log(n))) (-2^{-e^{-x arg(n)} cos(x log(n)))} - sin(log(2) e^{x arg(n)} sin(x log(n))) 2^{-e^{x arg(n)} cos( x log(n)))}\right)dx}$$

There may exist some other ways of evaluating the result, but this is an integral representation of it. Please correct me and give me feedback!

Тyma Gaidash
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