Calculating $\int_{0}^{0.5} \frac{\ln(1+x)}x$ with at least two decimal places precision. I'm preparing for a test and i've spent quite some time on this. What I already did was to use the taylor expansion for $\ln(1+x)$ to finally get the sum: $$ \sum_{n=1}^\infty \frac{(-1)^{n-1}}{{2^n} {n^2}} $$ I checked wolfram and it seems this gives the correct result, but how do I calculate this manually? I tried tweaking around with known sums, but to no avail. I also tried calculating the integral without the Taylor expansion approach, but it didn't work out. Any help will be greatly appreciated!
 A: Instead of writing a formula in summation notation of $\ln(1+x)$, I personally find it easier to expand the Taylor series expansion of $\ln (1+x)$, like this:
$$\ln (1+x)=x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\frac{x^5}5-\cdots,$$ where the expansion is sufficiently long enough to help you integrate and find your answer precisely to at least 2 decimal places. (The more terms in your series expansion, the more precise your answer will be. For me, I cut off after the fifth term upon integration.) 
Finally, 
\begin{align}
\int_0^{0.5} \frac{\ln(1+x)}x \, dx &\approx\int_0^{0.5} \frac{x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\frac{x^5}5}x \, dx \\
&= \int_0^{0.5}1-\frac x2+\frac{x^2}3-\frac{x^3}4+\frac{x^4}5 \, dx
\end{align}
Can you take it from here? Your answer should be very well precise to at least two decimal places. If you have more decimal places, simply truncate or round your answer accordingly.
A: Note that the terms in your series fall very rapidly to zero.  Since it is an alternating sum the error when summing up to and including term $N$ is bounded by the absolute value of term $N+1$.  You will not need many terms to get two digits precision.
A: By integrating termwise the Taylor series of $\frac{\log(1+x)}{x}$ around $x=0$ we get:
$$\int_{0}^{1/2}\frac{\log(1+x)}{x}\,dx = \sum_{j=1}^{+\infty}\frac{(-1)^{j+1}}{j^2\cdot2^j }$$
and through partial summation:
$$\sum_{j=1}^{+\infty}\frac{(-1)^{j+1}}{j^2\cdot 2^j }=\frac{1}{3}\sum_{j=1}^{+\infty}\frac{2j+1}{j^2(j+1)^2}+\frac{1}{3}\sum_{j=1}^{+\infty}\frac{(-1)^{j+1}(2j+1)}{2^j\cdot j^2(j+1)^2}=\frac{1}{3}+\frac{1}{3}\sum_{j=1}^{+\infty}\frac{(-1)^{j+1}(2j+1)}{2^j\cdot j^2(j+1)^2}$$
Performing partial summation a second time yields:
$$\sum_{j=1}^{+\infty}\frac{(-1)^{j+1}}{j^2\cdot 2^j }=\frac{1}{3}+\frac{1}{12}+\frac{1}{9}\sum_{j=1}^{+\infty}\frac{(-1)^{j+1}}{2^j}\left(\frac{1}{j^2}-\frac{2}{(j+1)^2}+\frac{1}{(j+2)^2}\right)$$
and performing the same trick a third time gives:
$$\sum_{j=1}^{+\infty}\frac{(-1)^{j+1}}{j^2\cdot 2^j }=\frac{427}{972}+\frac{1}{27}\sum_{j=1}^{+\infty}\frac{(-1)^{j+1}}{2^j}\left(\frac{1}{j^2}-\frac{3}{(j+1)^2}+\frac{3}{(j+2)^2}-\frac{1}{(j+3)^2}\right)$$
while the fourth time we have:
$$\sum_{j=1}^{+\infty}\frac{(-1)^{j+1}}{j^2\cdot 2^j }=\frac{1733}{3888}+\frac{1}{81}\sum_{j=1}^{+\infty}\frac{(-1)^{j+1}}{2^j}\sum_{k=0}^{4}\frac{(-1)^k\binom{4}{k}}{(j+k)^2}.$$
Finally, the first term in the RHS is $0.445\ldots$ while the first term in the last sum, multiplied by $\frac{1}{81}$, is less than $0.003$, hence $0.44$ is the approximation we were looking for.
The Euler acceleration method here slightly improves the Taylor series approximation since the error term goes to zero like $3^{-k}$ instead of $2^{-k}$.
