Finding the "Cauchy" $N,$ given $\varepsilon$, for the series $\sum n^{-3.5}$. We know that
$$\sum a_k \text{ converges} \iff \text{the partial sums } s_n \text{converge}
\iff \text{the partial sums } s_n \text{are Cauchy}$$ 
Writing out what this last statement means
$$\forall \varepsilon \gt 0, \exists N, \text{such that } \forall m \ge n \gt N, \left \lvert \sum_{k=n}^{m} a_k \right \rvert \lt \varepsilon$$
Let $\displaystyle a_k = \frac{1}{k ^{3.5}}$ and let $\displaystyle \varepsilon = 10^{−4}$. 
Find a value of N that satisfies the Cauchy
condition written out above.
 A: We are trying to make sure that
$$\sum_{k=a}^b \frac{1}{k^{3.5}}$$
is small.  It is OK to give away a whole lot. Note that if $k \ge N$, then $$k^{3.5}> N^{1.5}k(k-1).$$
It follows that if $a>N$, then 
$$\sum_{k=a}^b \frac{1}{k^{3.5}}< \sum_{k=a}^b \frac{1}{N^{1.5}}\frac{1}{k(k-1)}.$$
Note the partial fraction decomposition $\dfrac{1}{k(k-1)}=\dfrac{1}{k-1}-\dfrac{1}{k}$. So our sum has wholesale cancellation (telescoping), and 
$$\sum_{k=a}^b \frac{1}{k^{3.5}} <\frac{1}{N^{1.5}}\frac{1}{a-1}.$$
Since $a>N$, our sum is less than $\dfrac{1}{N^{2.5}}$
It is now easy to find $N$ such that ensures that ou sum is $<10^{-4}$.
Another way: By drawing a picture we can see that
$$\sum_{k=a}^b \frac{1}{k^{3.5}}<\int_{a-1}^\infty \frac{dx}{x^{3.5}}.$$
This integral is easily evaluated. If $a >N$, the integral is $\le \frac{1}{2.5}N^{-2.5}$.  This estimate is a better one than the one obtained earlier. But quality of the estimate is not really an issue, we want to prove only that there is an $N$ that does the job, and are not looking for the smallest $N$ that does. 
A: Hint 


*

*If you have positive terms and if you know it converges you may put try to find $n$ so that $$\sum_n^\infty a_k <\varepsilon\qquad \text{(why would this be sufficient?)}$$

*You might like to use some series where you know $\sum b_k$ and where $a_k\leq b_k$.
