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Find upper and lower bound for the following finite sum

$1/(1 + 1^3)+1/(1 +2^3)+1/(1 + 3^3) + ··· + 1/(1 + n^3)$

My attempt:

$1/(1 + 1^3)+1/(1 +2^3)+1/(1 + 3^3) + ··· + 1/(1 + n^3)$ = $\sum_{i=1}^n 1/(1+i^3)$ = $\int_1^n$1/$(1+i^3)$di = $\int_1^n1/(1+x^3)$dx

But now I'm stuck.Is my attempt correct?

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    $\begingroup$ It'a not correct. You can't substitute sum with an integral $\endgroup$ – Jakobian Feb 16 at 15:55
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No, because$$\sum_{i=1}^n\frac1{1+i^3}\not=\int_1^n\frac1{1+x^3}\,\mathrm dx.$$However,$$\int_i^{i+1}\frac1{1+x^3}\,\mathrm dx\leqslant\frac1{1+i^3}\leqslant\int_{i-1}^i\frac1{1+x^3}\,\mathrm dx,$$and you can use this to solve your problem.

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  • $\begingroup$ How can I find upper and lower bounds using this. I don't get it $\endgroup$ – dodo bc Feb 16 at 16:08
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    $\begingroup$ $$\int_0^n\frac1{1+x^3}\,\mathrm dx\leqslant\sum_{i=1}^n\frac1{1+i^3}\leqslant\int_1^{n+1}\frac1{1+x^3}\,\mathrm dx.$$ $\endgroup$ – José Carlos Santos Feb 16 at 16:11
  • $\begingroup$ so the lower bound is $\int_0^n {1/1+x^3}$ and the upper one is $\int_1^{n+1} {1/1+x^3}$ ? $\endgroup$ – dodo bc Feb 16 at 16:16
  • $\begingroup$ At least, those are the lower bounds that I got. $\endgroup$ – José Carlos Santos Feb 16 at 16:18
  • $\begingroup$ Which ones are the loower bounds? I'm losttt $\endgroup$ – dodo bc Feb 16 at 16:20

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