# Stuck on a problem about the length of a partial sum

I have a bit of a problem with this task since I couldn't really find any information in my books or so and I couldn't really find any similar to this problem on the internet so I hope someone doesn't mind giving me a hand. $$\sum_{k=1}^n \frac{(-1)^{k-1}}{k}$$ (where n goes to infinity) the problem is to find the length of the partial sum so that we get ${\log_{e}{2}}$ with exact value up to $10^{-2}$ .

• If I understand correctly, the question you should answer is "How large must $n$ be for the difference between the partial sum and the limit (which actually is $\log 2 = \ln 2$, and not $\log_{10} 2$) to be less than $10^{-2}$ in absolute value?" Jun 21, 2018 at 11:55
• I think that should be it,yes,I messed it up a bit. Jun 21, 2018 at 13:09

The alternating series theorem gives you a bound. For a series where the signs alternate and the terms decrease in magnitude the error is always of the sign of the first neglected term and smaller in magnitude. This guarantees that you will be within $10^{-2}$ after $100$ terms.
I don't know any way to find the minimum number of terms except by trying it. I made a spreadsheet and the error drops below $0.01$ at term $50$. For this series the terms are very slowly decreasing in absolute value.
I don't find it surprising that $\ln 2$ is very close to the average of two successive partial sums. That would suggest you want $\frac 12\cdot \frac 1n=0.01$, which gives $n=50$. This suggests we define $A(n)=\sum_{k=1}^n \frac{(-1)^{k-1}}{k}$ and take as our approximation $\frac 12(A(n)+A(n+1))$. That is within $0.01$ of $\ln 2$ at $n=4$ and converges much more quickly.
• The error does not drop to a minimum. It just gets smaller and smaller. I just made a table and looked where it dropped below $0.01$. Jun 21, 2018 at 14:47