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Using the integral test, find the best possible estimation for the series ∑n=1∞1n3\displaystyle\sum_{n=1}^\infty \frac{1}{n^3}n=1∑∞n31 using a partial sum with 101010 terms.
Note that ∑n=1101n3≈1.19753 \displaystyle\sum_{n=1}^{10} \frac{1}{n^3} \approx 1.19753 n=1∑10n31≈1.19753 and 1112≈0.00826\tfrac{1}{11^2} \approx 0.008261121≈0.00826.
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