# find the upper and lower bound for a finite sum

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?

• It'a not correct. You can't substitute sum with an integral – Jakobian Feb 16 at 15:55

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.
• $$\int_0^n\frac1{1+x^3}\,\mathrm dx\leqslant\sum_{i=1}^n\frac1{1+i^3}\leqslant\int_1^{n+1}\frac1{1+x^3}\,\mathrm dx.$$ – José Carlos Santos Feb 16 at 16:11
• so the lower bound is $\int_0^n {1/1+x^3}$ and the upper one is $\int_1^{n+1} {1/1+x^3}$ ? – dodo bc Feb 16 at 16:16