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.

Consider $\sum_{n=1}^{\infty} \frac{1}{n^{3}}$. We know that it converges.

Given $k\in\left(0,\infty\right)$. Is it then "okay" to say that there exists a $j\in\mathbb{N}$ such that $\sum_{n=j}^{\infty} \frac{1}{n^{3}}<k$? Or does it need some extra argumentation?

share|improve this question

2 Answers 2

up vote 2 down vote accepted

It always depends on the audience, but for an audience knowledgeable about sequences and series this should be obvious enough not to require further detailed arguments.

share|improve this answer
    
I agree since it is an elementary fact that a series converges only if its terms tend to $0$. –  robjohn Sep 26 '12 at 18:02

The fact $\sum\limits_{n=1}^{\infty} \frac{1}{n^{3}}$ that converges means that converges the sequence of partial sums $S_N=\sum\limits _{n=1}^{N} \frac{1}{n^{3}}$

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.