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Is there a formula for the sum of the series $\displaystyle \sum_{i=0}^N r^{1/(a+bi)}$ similar to $\displaystyle \sum_{i=0}^N r^i =\frac{1-r^N}{1-r}$ ?

$a$ and $b$ are constants. Thanks for your help.

Jani
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  • is $i$ the imaginary unit? – Dr. Sonnhard Graubner Jun 01 '17 at 18:27
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    @Dr.SonnhardGraubner It is the summation index, based on the question. – Clement C. Jun 01 '17 at 18:29
  • I doubt there is a formula. For most finite sums there is no simple formula - the geometric sum is something of an exception to the general rule. – Jair Taylor Jun 01 '17 at 18:33
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    I agree with Clement although the $a+bi$ was suspicious. I almost changed the summation index to $k$ in my edits but figured there was a small chance OP did mean complex numbers and just improperly used notation. It's happened before on this site. The waffling made me lose interest.. eh. –  Jun 01 '17 at 18:33

1 Answers1

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Just added for your curiosity.

As already written in comments, I do not think that there exists any closed form formula for $$S_n=\sum_{i=0}^n r^\frac{1}{a+bi}$$ However, we could $\color{red}{\text{approximate}}$ them considering $$I=\int r^\frac{1}{a+bx}\,dx=\int\exp\left(\frac{\log (r)}{a+b x} \right)\,dx=\frac 1b\int \exp\left(\frac{\log (r)}{y} \right)\,dy$$ which is $$I=\frac 1b\left(y\, r^{\frac{1}{y}}-\log (r)\,\text{Ei}\left(\frac{\log (r)}{y}\right)\right)$$ where appears the exponential integral function.

Back to $x$ and using the bounds $$T_n=\int_0^n r^\frac{1}{a+bx}\,dx$$ write $$T_n=\frac 1b\left(\log (r) \left(\text{Ei}\left(\frac{\log (r)}{a}\right)-\text{Ei}\left(\frac{\log (r)}{a+b\,n}\right)\right)+(a+b\, n) \, r^{\frac{1}{a+b n}}-a\, r^{\frac{1}{a}}\right)$$ For illustration purposes, I give below a table for $r=\frac 12$ and for a few values of $a,b$ $$\left( \begin{array}{ccccc} a & b & n & T_n & S_n \\ 1 & 1 & 10 & 9.22372 & 8.53167 \\ 1 & 1 & 20 & 18.7997 & 18.0937 \\ 1 & 1 & 30 & 28.5385 & 27.8274 \\ 1 & 1 & 40 & 38.3493 & 37.6355 \\ 1 & 1 & 50 & 48.2008 & 47.4853 \\ & & & & \\ 1 & 2 & 10 & 9.72945 & 9.04685 \\ 1 & 2 & 20 & 19.5080 & 18.8177 \\ 1 & 2 & 30 & 29.3740 & 28.6810 \\ 1 & 2 & 40 & 39.2776 & 38.5832 \\ 1 & 2 & 50 & 49.2022 & 48.5070\\ & & & & \\ 1 & 3 & 10 & 9.94459 & 9.27580 \\ 1 & 3 & 20 & 19.7947 & 19.1207 \\ 1 & 3 & 30 & 29.7046 & 29.0287 \\ 1 & 3 & 40 & 39.6399 & 38.9631 \\ 1 & 3 & 50 & 49.5894 & 48.9120\\ & & & & \\ 2 & 1 & 10 & 9.66759 & 8.85178 \\ 2 & 1 & 20 & 19.2687 & 18.4406 \\ 2 & 1 & 30 & 29.0171 & 28.1843 \\ 2 & 1 & 40 & 38.8329 & 37.9975 \\ 2 & 1 & 50 & 48.6875 & 47.8506\\ & & & & \\ 2 & 2 & 10 & 10.0393 & 9.22030 \\ 2 & 2 & 20 & 19.8249 & 18.9988 \\ 2 & 2 & 30 & 29.6934 & 28.8647 \\ 2 & 2 & 40 & 39.5983 & 38.7683 \\ 2 & 2 & 50 & 49.5238 & 48.6929\\ & & & & \\ 2 & 3 & 10 & 10.2102 & 9.39475 \\ 2 & 3 & 20 & 20.0637 & 19.2431 \\ 2 & 3 & 30 & 29.9747 & 29.1524 \\ 2 & 3 & 40 & 39.9106 & 39.0874 \\ 2 & 3 & 50 & 49.8605 & 49.0367\\ & & & & \\ 3 & 1 & 10 & 9.90856 & 9.04240 \\ 3 & 1 & 20 & 19.5319 & 18.6549 \\ 3 & 1 & 30 & 29.2892 & 28.4078 \\ 3 & 1 & 40 & 39.1098 & 38.2260 \\ 3 & 1 & 50 & 48.9674 & 48.0821\\ & & & & \\ 3 & 2 & 10 & 10.1998 & 9.32743 \\ 3 & 2 & 20 & 19.9920 & 19.1130 \\ 3 & 2 & 30 & 29.8630 & 28.9815 \\ 3 & 2 & 40 & 39.7692 & 38.8864 \\ 3 & 2 & 50 & 49.6955 & 48.8119\\ & & & & \\ 3 & 3 & 10 & 10.3415 & 9.46925 \\ 3 & 3 & 20 & 20.1981 & 19.3210 \\ 3 & 3 & 30 & 30.1102 & 29.2314 \\ 3 & 3 & 40 & 40.0467 & 39.1670 \\ 3 & 3 & 50 & 49.9970 & 49.1167 \end{array} \right)$$

Claude Leibovici
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