How to estimate the sum of the alternate series to within 0.1 of its true value?
I am given $\sum_{n=1} \frac{(-1)^{n+1}}{n}$. It is understood that the sequence converges but I am specifically looking for how to set up and use Leibniz' estimate to find the sum.
For alternating series $\sum_n (-1)^{n+1} a_n$ where $(a_n)$ is a decreasing sequence convergent to $0$ we have
$$\vert R_n\vert:=\left\vert\sum_{k=n+1}^\infty (-1)^{k+1}a_k\right\vert\le a_{n+1}$$ so in your case you need to find $n$ such that $$\frac1{n+1}\le 0.1$$ and so an estimate of the sum is $\sum_{k=1}^n\frac{(-1)^{k+1}}k$ with precision $0.1$.
