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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.

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  • $\begingroup$ 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
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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 $a_{n+1}$.

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

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