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!


1 Answer 1


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 $$ m_{\rm best}={(20000)^{100}}\approx 1.26765 10^{430}. $$


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