Calculus Help!!
By recognizing the series
-((1/4))-((1/4))^2/2-((1/4))^3/3-...-(…
as a Taylor series evaluated at a particular value of x, find the sum of the convergent series.

Question

anonymous · Answer

$$
\frac 1 {1-x} = \sum_i^\infty x^n, \quad |x| <1 

$$
$$
\int_0^x \frac 1 {1-t}dt = \sum_{i=0}^\infty \int_0^x t^n dt, \quad |x| <1 

$$

$$
-\ln(1-x) = \sum_{i=0}^\infty \frac {x^{n+1}}{n+1}, \quad |x| <1 

$$

It is easy from now on.

anonymous · Answer

-ln(1 - 1/4) = -ln(3/4) = ln(4/3)

anonymous · Answer

and i put ln(4/3) and it is incorrect!!

dumbcow · Answer

try ln(3/4) since the sum should be a neg number
$$\rightarrow -\sum_{0}^{\infty}\frac{x^{n+1}}{n+1} = \ln(1-x)$$

anonymous · Answer

it is corrrect!!Thanks, @dumbcow =))

anonymous · Answer

Thanks for helping @eliassaab!!

dumbcow · Answer

welcome...eliassaab did the work though :)