Real Analysis II --- power series If f(x)= $\int_0^x \frac {ln(1+u)}{u}$ du, find a series for f(x) and calculate the approximate value of f(0.1). Use the error upper bound for approximating an alternating series to give an upper bound for the error in your approximation. It is up to you to decide how precise your estimate will be. 
Attempt:
Given ln(1+x)=$\sum_{n=1}^\infty$ $(-1)^{n+1} \frac {u^n}{n}$, we have $\frac{(-x)^{n-1}}{n}$=$\frac {ln(1+u)}{u}$. So f(x)=$\sum_{n=1}^\infty$ $(-1)^{n-1} \frac {u^n}{n^2}$ and then I just plug u=.1 and add terms together to n= a large enough number I deem precise enough?
 A: As Ron Gordon commented, (I typed this before your edit), start with $$\log(1+u)=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{u^n}n$$ so, $$\frac {ln(1+u)}{u}=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{u^{n-1}}n$$ $$\int\frac {ln(1+u)}{u}du=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{u^{n}}{n^2}$$ $$f(x)=\int_0^x\frac {ln(1+u)}{u}du=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^{n}}{n^2}$$ $$f(\frac 1 {10})=\frac{1}{10}-\frac{1}{400}+\frac{1}{9000}-\frac{1}{160000}+\frac{1}{2500000}+\cdots$$ Just think about Lagrange remainder for a Taylor  series (in your case, what is interesting is that the series alternates, so $\cdots$).
Edit
As said in a comment, we can explicitly know the number of terms $ k$ to be added for a given accuracy $\epsilon$. Using Lagrange remainder, we then need to solve $$\frac{x^k}{n^2} \leq \epsilon $$ Any equation which can write $A+B x+C\log(D+Ex)=0$ has an explicit solution in terms of Lambert function.
In the case of the current problem the solution is then given by $$n \geq-\frac{2}{\log (x)}W\left(\frac{\log ^2(x) \sqrt{\frac{\epsilon }{\log ^2(x)}}}{2 \epsilon }\right)=-\frac{2}{\log (x)}W\left(\frac{|\log (x)|}{2 \sqrt{\epsilon }}\right)$$ What is nice is that there are very good approximation of Lambert function $W(z)$. The first terms are given by $$W(z)\approx L_1-L_2+\frac{L_2}{L_1}+\frac{L_2(L_2-2)}{2L_1^2}+\cdots$$ where $L_1=\log(z)$, $L_2=\log(L_1)$.
For illustration purposes, let us take $x=\frac 1{10}$ and $\epsilon=10^{-6}$; these make the argument of Lambert function $z=500 \log (10)$ so the approximation gives $W(500 \log (10))\approx 5.37193$ (the exact value being $\approx 5.36816$) and, so, $n \geq 4.66272$. So, five terms are required.
Let us check (easy, since these are the terms I wrote). Using five terms, we get $$f(\frac 1 {10})=\frac{17568947}{180000000}\approx 0.09760526111$$ while the exact value of the integral would be $$\int_0^{\frac 1 {10}} \frac {ln(1+u)}{u}du=-\text{Li}_2\left(-\frac{1}{10}\right)\approx 0.09760523523$$ 
Please, have a detailed look at Ron Gordon's comment
