Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

2 Answers 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.

share|improve this answer

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 $a_k\leq b_k$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.