a convergent series with x<1, and a, b are constants:

$\sum_{i=a}^{inf} \frac{x^i}{1+ i \cdot b}$

what is and how to find its closed form ???

2. Originally Posted by graticcio
a convergent series with x<1, and a, b are constants:

$\sum_{i=a}^{inf} \frac{x^i}{1+ i \cdot b}$

what is and how to find its closed form ???

I am going to assume $a=0$. Say that $b=1$.

We want to sum, $\sum_{k=0}^{\infty} \frac{x^k}{k+1}$

Consider $\sum_{k=0}^{\infty} x^k = \frac{1}{1 - x}$.

That means, $\int_0^t \sum_{k=0}^{\infty} x^{k} dx = \int_0^t \frac{1}{1-x} dx$

Thus, $\sum_{k=0}^{\infty} \int_0^t x^{k} dt = - \ln (1-t)$ for $0

Thus, $\sum_{k=0}^{\infty} \frac{t^{k+1}}{k+1} = - \ln (1-t)$

Thus, $\sum_{k=0}^{\infty} \frac{t^k}{k+1} = - t^{-1} \ln (1-t)$

----
In the general case consider, $\sum_{k=0}^{\infty} x^{bk} = \frac{1}{1-x^b}$
And apply a similar argument.

i hav tried a similar approach
but when $a \neq 0$

$\frac{x^{ab}}{1 - x^b}$

and the integration of it
$\int \frac{x^{ab}}{1 - x^b} dx$
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

4. Originally Posted by graticcio
a convergent series with x<1, and a, b are constants:

$\sum_{i=0}^{\infty} \frac{x^i}{1+ i \cdot b}$

what is and how to find its closed form ???
$\sum\limits_{i = 0}^\infty {\frac{{x^i }}
{{1 + bi}}} = \sum\limits_{i = 0}^\infty {x^i \int_0^1 {u^{bi} \,du} } = \int_0^1 {\left[ {\sum\limits_{i = 0}^\infty {\left( {xu^b } \right)^i } } \right]\,du} = \int_0^1 {\frac{1}
{{1 - xu^b }}\,du} .$

There's your closed form.

5. thanks

what if the series does not start with 0
but some constant $a$ ???

6. Originally Posted by graticcio
a convergent series with x<1, and a, b are constants:

$\sum_{i=a}^{inf} \frac{x^i}{1+ i \cdot b}$

what is and how to find its closed form ???

As an entree:

For the special case a = 0 and b = 1, it's not hard to use the standard result for an infinite geometric series to show that the series is equal to $-\frac{\ln (1 - x)}{x}$.

7. Originally Posted by mr fantastic
As an entree:

For the special case a = 0 and b = 1, it's not hard to use the standard result for an infinite geometric series to show that the series is equal to $-\frac{\ln (1 - x)}{x}$.
But I see Krizalid's skipped the soup du jour and gone to the main course. That just leaves desert:

Originally Posted by graticcio
thanks

what if the series does not start with 0
but some constant $a$ ???
Then you just relabel the dummy index ......

$\sum_{i=a}^{inf} \frac{x^i}{1+ i \cdot b} = \sum_{n=0}^{inf} \frac{x^{n+a}}{1+ (n+a) \cdot b} = .....$

8. thanks

yes, shifting the index, but then it comes back
to the problem i mentioned in the second reply

$\sum\limits_{n = 0}^\infty {\frac{{x^{n+a}}} {{1 + b(n+a)}}}= \int_0^1 {\left[ {\sum\limits_{n = 0}^\infty {\left( {xu^b } \right)^{n+a} } } \right]\,du} = \int_0^1 {\frac{(xu^b)^a} {{1 - xu^b }}\,du}$

cannot find the 'closed' form of that integral

9. Originally Posted by graticcio
thanks

yes, shifting the index, but then it comes back
to the problem i mentioned in the second reply

$\sum\limits_{n = 0}^\infty {\frac{{x^{n+a}}} {{1 + b(n+a)}}}= \int_0^1 {\left[ {\sum\limits_{n = 0}^\infty {\left( {xu^b } \right)^{n+a} } } \right]\,du} = \int_0^1 {\frac{(xu^b)^a} {{1 - xu^b }}\,du}$

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.

10. thanks