by phw » Sat Aug 11, 2018 10:37 pm
I doubt there is a simple closed form for that sum. Will you accept an approximation?
First subtract 1 from each term and add those separately.
[tex]\sum_{k=1}^{1000 }1 + \sum_{k=1}^{1000 }\frac{2}{2k-1}[/tex]
The first term is just 1000. The second term is the sum of two well known series, the harmonic series [tex]H(2000) = 1 + \frac{1}{2}+ \frac{1}{3}+\frac{1}{4}+\frac{1}{5}+ ...+\frac{1}{1999}+ \frac{1}{2000}[/tex]
and the alternating series [tex]A(2000) = 1 - \frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5} ... +\frac{1}{1999}- \frac{1}{2000}[/tex].
The harmonic series is well approximated by [tex]H(n) =\ln n + \gamma[/tex] where [tex]\gamma \approx0.577[/tex] is Euler's constant.
The alternating harmonic series converges to [tex]\ln 2[/tex]. It converges slowly, but with 2000 terms the approximation is good for 3 decimals.
So your sum is about [tex]1000+ \ln 2000 + 0.577 + \ln 2[/tex].