For the series below calculate the sum of the first 3 terms, S3, and find a bound for the error.

$$ \sum_{n=1}^\infty\frac{200(-1)^n}{n^{0.7}} $$

For the first three terms I got 381.586.. S_3, not sure if it's right.

For finding the with an $|\text{error}|<=$, I have no clue.

  • $\begingroup$ It does not look right to me. You should have about $-200 +123.11444 -92.69261 \approx -169.5782$. The full sum should be between that and $0$ and is in fact about $-128.44$ $\endgroup$ – Henry Nov 24 '16 at 23:30
  • $\begingroup$ @Henry yeah figured this out, having issues with the error bound, ty $\endgroup$ – Robert Miller Nov 24 '16 at 23:52

For an alternating series $\sum (-1)^ka_k$, $(a_k)_k$ positive, monotonously falling to zero, the Leibniz test also provides an error bound. For $s_n$ the error bound is the next term $a_{n+1}$.

| cite | improve this answer | |
  • $\begingroup$ Still don't know how to approach this. $\endgroup$ – Robert Miller Nov 25 '16 at 0:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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