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.

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Could compare the sums to (simpler to evaluate) integrals. –  Did Mar 8 '12 at 6:46
@max What do you know about $\sum k^{-2}$? What do you now about $k^{-3.5}$ vs $k^{-2}$ when $k>1$?. What does this tell you about the ultimate value of your sum? –  Pedro Tamaroff Mar 8 '12 at 6:54
You can make us of Euler-Maclaurin summation to come with really tight error bounds. en.wikipedia.org/wiki/Euler-Maclaurin_formula –  user17762 May 8 '12 at 3:48

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.

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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 $a_k\leq b_k$.

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