<s>[INST] Trying to show $1-\frac12 -\frac {1} {4}+\frac {1} {3}-\frac {1} {6}-\frac {1} {8}+\frac {1} {5}-\cdots =\frac {1} {2}\log 2$ Now I think the lhs can be rewritten as $$\sum _{n=1}^\infty \left( \dfrac {1} {2n-1}-\dfrac {1} {2\cdot 3^{n-1}}-\dfrac {1} {2^{n+1}}\right) =\dfrac {1} {2}\log 2$$
I guess one way to do this may be to start from RHS and try to bring up an expression which is same as the LHS.
so using log series expansion $$\frac {1} {2}\log 2 = \log \sqrt {2} = \sum _{n=1}^\infty \dfrac {\left( -1\right) ^{n+1}} {n}\left( \sqrt {2}-1\right) ^{n}$$
Now Binomial expansion Theorem seems to be making a calling here.
$$\sum _{n=1}^{\infty }\dfrac {\left( -1\right) ^{n+1}} {n}\left( \sqrt {2}-1\right) ^{n} = \sum _{n=1}^{\infty }\dfrac {\left( -1\right) ^{n+1}} {n}\sum _{k=0}^{k=n}C_{k}^{n} \sqrt {2}^{k}\left( -1\right) ^{n-k} $$
I am unsure how to proceed from here and even if i am going down the right path.
Any help would be much appreciated.
 [/INST] Rewrite $1-\dfrac {1} {2} -\dfrac {1} {4}+\dfrac {1} {3}-\dfrac {1} {6}-\dfrac {1} {8}+\dfrac {1} {5}-\cdots$ as 
$$\left(1-\dfrac {1} {2}\right) -\dfrac {1} {4}+\left(\dfrac {1} {3}-\dfrac {1} {6}\right)-\dfrac {1} {8}+\left(\dfrac {1} {5} - \dfrac{1}{10}\right) - \dfrac{1}{12} \ldots$$
or
$$\frac{1}{2} - \frac{1}{4} + \frac{1}{6} - \frac{1}{8} + \frac{1}{10} - \frac{1}{12} + \cdots = \frac{1}{2}\left(\frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots\right) = \frac{1}{2} \log(2).$$
Edit: More generally (without numbers), we write the left hand side as $\displaystyle\sum_{k=1}^\infty  \frac{1}{2k-1} - \frac{1}{4k-2} - \frac{1}{4k} $ which is equal to $\displaystyle\sum_{k=1}^\infty  \frac{1}{4k-2}- \frac{1}{4k} $ (since $\dfrac{1}{2k-1} - \dfrac{1}{4k-2} = \dfrac{1}{2k-1} - \left(\dfrac{1}{2k-1}\right)\dfrac{1}{2} =  \left(\dfrac{1}{2k-1}\right)\dfrac{1}{2}$).
This is equal to $\left(\dfrac{1}{2}\right)\displaystyle\sum_{k=1}^\infty  \frac{1}{2k-1}- \frac{1}{2k} $ $= \dfrac{1}{2}\displaystyle\sum_{k=1}^\infty \dfrac{(-1)^{k+1}}{k}$, or $\dfrac{\log(2)}{2}$ by examination of the power series of $\log$.
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