Arithmetic and Geometric progressions.


Postby Guest » Fri Aug 10, 2018 2:40 am

Find the sum of the following progression from 1st term to 1000th term.
3, 5/3, 7/5, 9/7, 11/9, 13/11……..
Also, what is the type of the progression ?

Re: Progressions

Postby 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].

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