a convergent series with x<1, and a, b are constants:
$\displaystyle \sum_{i=a}^{inf} \frac{x^i}{1+ i \cdot b}$
what is and how to find its closed form ???
please help, many thanks!
I am going to assume $\displaystyle a=0$. Say that $\displaystyle b=1$.
We want to sum, $\displaystyle \sum_{k=0}^{\infty} \frac{x^k}{k+1}$
Consider $\displaystyle \sum_{k=0}^{\infty} x^k = \frac{1}{1 - x}$.
That means, $\displaystyle \int_0^t \sum_{k=0}^{\infty} x^{k} dx = \int_0^t \frac{1}{1-x} dx$
Thus, $\displaystyle \sum_{k=0}^{\infty} \int_0^t x^{k} dt = - \ln (1-t)$ for $\displaystyle 0<t<1$
Thus, $\displaystyle \sum_{k=0}^{\infty} \frac{t^{k+1}}{k+1} = - \ln (1-t)$
Thus, $\displaystyle \sum_{k=0}^{\infty} \frac{t^k}{k+1} = - t^{-1} \ln (1-t)$
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In the general case consider, $\displaystyle \sum_{k=0}^{\infty} x^{bk} = \frac{1}{1-x^b}$
And apply a similar argument.
thanks for your reply
i hav tried a similar approach
but when $\displaystyle a \neq 0$
we hav to start with
$\displaystyle \frac{x^{ab}}{1 - x^b}$
and the integration of it
$\displaystyle \int \frac{x^{ab}}{1 - x^b} dx$
i hav checked with some integration tables
but there is only reduction formula available
i wonder how to get a "more closed" form of the integral
or if there is any other way to find the sum of tht series
$\displaystyle \sum\limits_{i = 0}^\infty {\frac{{x^i }}
{{1 + bi}}} = \sum\limits_{i = 0}^\infty {x^i \int_0^1 {u^{bi} \,du} } = \int_0^1 {\left[ {\sum\limits_{i = 0}^\infty {\left( {xu^b } \right)^i } } \right]\,du} = \int_0^1 {\frac{1}
{{1 - xu^b }}\,du} .$
There's your closed form.
But I see Krizalid's skipped the soup du jour and gone to the main course. That just leaves desert:
Then you just relabel the dummy index ......
$\displaystyle \sum_{i=a}^{inf} \frac{x^i}{1+ i \cdot b} = \sum_{n=0}^{inf} \frac{x^{n+a}}{1+ (n+a) \cdot b} = .....$
thanks
yes, shifting the index, but then it comes back
to the problem i mentioned in the second reply
$\displaystyle \sum\limits_{n = 0}^\infty {\frac{{x^{n+a}}} {{1 + b(n+a)}}}= \int_0^1 {\left[ {\sum\limits_{n = 0}^\infty {\left( {xu^b } \right)^{n+a} } } \right]\,du} = \int_0^1 {\frac{(xu^b)^a} {{1 - xu^b }}\,du} $
cannot find the 'closed' form of that integral
If you enter the integrand into the Wolfram integrator (make sure you use x, not u as your integration variable), you'll get a closed form for the indefinite integral. Now evaluate the definite integral.