Calculate $\sum_{k=51}^{\infty} \frac{(-1)^{k-1}}{k} $ I have a sum to calculate:
$$\sum_{k=51}^{\infty} \frac{(-1)^{k-1}}{k} $$
And I have no idea about how to proceed. What kind of techniques are available to calculate this?
 A: Start with the Taylor series of $\log(1+x)$:
$$\log(1+x) = \sum_{k=1}^\infty\frac{(-1)^{k-1}x^k}{k}$$
and take $x=1$ (that this is allowed follows from Abel's theorem) to find
$$\log(2) = \sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}$$
Now split the sum into two parts:
$$\log 2 = \sum_{k=1}^\infty\frac{(-1)^{k-1}}{k} = \sum_{k=1}^{50}\frac{(-1)^{k-1}}{k} + \sum_{k=51}^\infty\frac{(-1)^{k-1}}{k}$$
The first term can be calculated on a computer (there is no known (useful) formula you can use so a calculation is needed here)
$$ \sum_{k=1}^{50}\frac{(-1)^{k-1}}{k} = \frac{2117413358024481287753}{3099044504245996706400} \approx 0.68324716$$
Thus giving us
$$\sum_{k=51}^\infty\frac{(-1)^{k-1}}{k} = \log 2 - \frac{2117413358024481287753}{3099044504245996706400} \approx 0.00990002$$
If you are only looking for an approximate value: The terms in the series $\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}$ alternates being positive and negative and the absolute value of the terms decreases with $k$. This means that the sum you are after satsify:
$$\left|\sum_{k=51}^\infty\frac{(-1)^{k-1}}{k}\right| = \left|\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k} - \sum_{k=1}^{50}\frac{(-1)^{k-1}}{k}\right| < \left|\frac{(-1)^{51-1}}{51}\right| = \frac{1}{51}$$
so the sum you are after is less than $0.02$ (which is indeed the case).
