2
$\begingroup$

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.

  • 0
    You can make us of Euler-Maclaurin summation to come with really tight error bounds. http://en.wikipedia.org/wiki/Euler-Maclaurin_formula2012-05-08

2 Answers 2

2

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.

0

Hint

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

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