Hi I stumbled across an alternating harmonic sum with a fixed limit on an practice exam and I've no idea how to calculate this sum.
\begin{equation}
\sum_{1}^{100}\frac{(-1)^{k-1}}{k}. \tag{1}
\end{equation}
Now the point of the exercise is to make an approximation to this sum and compare it to
$$\sum_{1}^{\infty}\frac{(-1)^{k-1}}{k} \tag{2}.$$
So naturally I tried to convert $(1)$ into an integral so I could make an approximation based on that, however I found it hard to integrate $$\frac{\cos(x)}{x}. $$
Then I tried to make use of Leibniz criterion but that got me nowhere. Since I guess there is a way to make an approximation of $(1)$. I would gladly accept any tip on how to solve this!
Thanks for reading! :)
Answer
Starting from par's answer, you can use the expansion of the harmonic numbers for large values and, in such a case, you find that, for even values of $n$, $$\begin{equation}
S=\sum_{k=1}^{2m}\frac{(-1)^{k-1}}{k}=\log (2)-\frac{1}{4 m}+\frac{1}{16 m^2}-\frac{1}{128
m^4}+O\left(\left(\frac{1}{m}\right)^5\right)
\end{equation}$$ Using this expansion, for $m=50$, you will find $$S\approx \log (2)-\frac{3980001}{800000000} =0.68817217930994530942$$ while the exact value would be $$S=\frac{47979622564155786918478609039662898122617}{697203752297124771645338089353123035
56800} \approx 0.68817217931019520324$$
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