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Question Number 74885 by abdomathmax last updated on 03/Dec/19

calcilate Σ_(n=1) ^(16)  (1/n^3 )

$${calcilate}\:\sum_{{n}=\mathrm{1}} ^{\mathrm{16}} \:\frac{\mathrm{1}}{{n}^{\mathrm{3}} } \\ $$

Answered by tw000001 last updated on 05/Dec/19

I use integral to solve.  Σ_(n=1) ^(16) (1/n^3 )=∫_(16) ^(17) Σ_(n=1) ^n (1/n^3 )dn  =∫_1 ^(17) Σ_(n=1) ^n (1/n^3 )dn−∫_1 ^(16) Σ_(n=1) ^n (1/n^3 )dn≈1.2002

$$\mathrm{I}\:\mathrm{use}\:\mathrm{integral}\:\mathrm{to}\:\mathrm{solve}. \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\mathrm{16}} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{3}} }=\int_{\mathrm{16}} ^{\mathrm{17}} \underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{3}} }{dn} \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{17}} \underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{3}} }{dn}−\int_{\mathrm{1}} ^{\mathrm{16}} \underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{3}} }{dn}\approx\mathrm{1}.\mathrm{2002} \\ $$

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