$$\sum_{n=1}^{\infty}\frac{14}{n^2+n}$$ Determine how big n has to be such that the error is less than 0.001.
I tried approximating the error using $|S-S_n^*|$, where $S\approx S_{n}^{*}= S_n +\frac{A_{n+1}+A_n}{2}, A_n=\int_n^{\infty}f(x)dx$ so that it satisfies $|S-S_n^*|\le\frac{A_n-A_{n+1}}{2}$. I got that $n\ge 13999.0002$ which means that $n$ has to be $14000$.. This doesn't seem to be correct, did I misunderstand the approximation method?
$\frac {14}{n^2 + n} = \frac {14}{n(n + 1)} = \frac {14}{n} - \frac {14}{(n + 1)}$
$\sum_\limits{n=1}^{\infty}\frac {14}{n^2 + n} = 14$
But that doesn't seem to be your question.
$\sum_\limits{n=1}^{\infty}\frac {14}{n^2 + n}=\sum_\limits{n=1}^{N-1} \frac {14}{n^2 + n} + \sum_\limits{n=N}^{\infty}\frac {14}{n^2 + n}$
and how big does N need to be for:
$\sum_\limits{n=N}^{\infty}\frac {14}{n^2 + n} < 0.001$
$N = \frac {14}{0.001} = 14,000$
But, if you want to estimate that using the integral $\int_N^{\infty} \frac {14}{x^2+x} dx$
$14 \ln (\frac {N}{N+1}) < 0.001$
And you still get $N\approx 14,000$