For the following series, find the number of terms required to find the sum with error < 0.005, and find upper and lower bounds for the sum using a much smaller number of terms.
$$\sum_{n=1}^\infty \frac 1{n^{1.01}}.$$ I really am not sure how to approach this problem - I've never seen anything like it. Help would be much appreciated! (I'm also pretty sure I've formatted it wrong, so feel free to correct me).
Please help me!! I'm pretty lost!
Let $s=1/100$. The error on considering the first $m$ terms is $$E_m=\sum_{n=m+1}^\infty\frac{1}{n^{1+s}}.$$ For $t\in[n,n+1]$ we have $$ \frac{1}{(n+1)^{1+s}}<\int_n^{n+1}\frac{dt}{t^{1+s}}<\frac{1}{n^{1+s}}$$ Thus, for $m>0$, $$ \int_{m+1}^\infty\frac{dt}{t^{1+s}}<E_m<\int_m^\infty\frac{dt}{t^{1+s}}. $$ Or $$\frac{1}{s(m+1)^s} <E_m< \frac{1}{s m^s} $$ It follows that $$ m < \left(sE_m\right)^{-1/s}<m+1. $$ This proves that the least $m$ that yields $E_m=\epsilon$ is $m_{\rm best}=\lceil{(s\epsilon)^{-1/s}}\rceil$. In our case $\epsilon=5/1000$ and $s=100$, so $$ m_{\rm best}={(20000)^{100}}\approx 1.26765 10^{430}. $$