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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.

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  • $\begingroup$ 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
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For 1dp the bound should be.05, not .01.

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