The series is $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}k}{k^2+1}$

The minimum number of terms needed should be given by $\lvert a_{n+1} \rvert$

Thus $\lvert \frac{(-1)^{k+2}(k+1)}{(k+1)^2+1} \rvert<0.01$

However, this is only satisfied after the $99th$ term, whereas the answer says the $19th$ term.

  • $\begingroup$ You could sum 100 terms that are around $0.009$ which matches your criterion, but that will still affect the decimal place. This will have something more to do with the rate that the series terms get smaller. $\endgroup$ – Benedict W. J. Irwin Jun 30 '18 at 10:34
  • $\begingroup$ Yes, now that I think about it, the answer is still wrong in a sense that it will still be affecting that decimal point after 19 terms, however, at 19 terms exactly it is correct to 1DP. I believe, there is another method to solve this rather than using the truncation error, but I am not quite sure. $\endgroup$ – Anthony P Jun 30 '18 at 10:38
  • $\begingroup$ Since the series is alternating, solve for $k$, $\frac{k+1}{(k+1)^2+1}=0.05$ (don't miss marty cohen's answer). $\endgroup$ – Claude Leibovici Jul 1 '18 at 14:07

For 1dp the bound should be.05, not .01.


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.