# find the precise range of of the following sum

I need to find the precise range of the following sum to infinity:

$$\sum_{n=1}^\infty\frac{x^n}{4^nn}$$

Can someone post how to do it because I have no idea, I'm also given that it converges...

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Do you want the range of the function defined by this sum (when it converges), or do you want the interval of convergence? I suspect it is the second. – André Nicolas Nov 21 '12 at 19:43

$$f(x) = \sum_{n=1}^{\infty} \dfrac{(x/4)^n}n$$ Consider $g(x) = \displaystyle \sum_{n=0}^{\infty} (x/4)^n$. This converges for $\vert x \vert < 4$ and it converges to $$\dfrac1{1-x/4} = \dfrac{4}{4-x}$$ Note that $$\dfrac{df}{dx} = \sum_{n=1}^{\infty} \dfrac{nx^{n-1}}{n4^n} = \dfrac14 \cdot \sum_{n=0}^{\infty} (x/4)^n = \dfrac14 g(x) = \dfrac1{4-x}$$ (You may want to read why term by term differentiation or integration of a power series within its domain of convergence is valid)

Hence, $$f(x) = \int_0^x \dfrac{dt}{4-t} + f(0)$$ And we know that $f(0) = 0$. Hence, $$f(x) = \int_0^x \dfrac{dt}{4-t} = - \log(4-t) \vert_{0}^x = \log 4 - \log(4-x)$$ Hence, $\displaystyle f(x) = \sum_{n=1}^{\infty} \dfrac{(x/4)^n}n$ converges for $\vert x \vert < 4$ and converges to $\log 4 - \log(4-x)$

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Let $$f(x)=\sum_{n\ge 1}\frac{x^n}{4^nn}=\sum_{n\ge 1}\frac1n\left(\frac{x}4\right)^n\;,$$ which is easily seen to converge for $-4\le x<4$. Then

$$f\,'(x)=\sum_{n\ge 1}\frac{x^{n-1}}{4^n}=\frac14\sum_{n\ge 0}\left(\frac{x}4\right)^n=\frac14\cdot\frac1{1-x/4}=\frac1{4-x}\;.$$

Now you know exactly what $f\,'(x)$ is, and you know that $f(0)=0$; can you use this information to find a nice closed expression for $f(x)$ and then use that to answer the question?

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The thing is I'm not sure what the answer as Im asked to specifie precise range of this sum and is that log 4 - log (4-x) or is it just -4 <= x <4 ? – mc1 Nov 21 '12 at 20:34
@mc1: $-4\le x<4$ is the interval of convergence of the sum. If that’s what is wanted, someone is using the expression range of the sum in a very sloppy way, but it’s possible. The sum is a function of $x$, and by range of the sum I would understand the range of that function, i.e., the set of all possible values of $f(x)$ for $x\in[-4,4)$. – Brian M. Scott Nov 21 '12 at 20:38