0
$\begingroup$

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?

|cite|improve this question
$\endgroup$
  • 1
    $\begingroup$ It'a not correct. You can't substitute sum with an integral $\endgroup$ – Jakobian Feb 16 at 15:55
1
$\begingroup$

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.

enter image description here

|cite|improve this answer
$\endgroup$
  • $\begingroup$ How can I find upper and lower bounds using this. I don't get it $\endgroup$ – dodo bc Feb 16 at 16:08
  • 1
    $\begingroup$ $$\int_0^n\frac1{1+x^3}\,\mathrm dx\leqslant\sum_{i=1}^n\frac1{1+i^3}\leqslant\int_1^{n+1}\frac1{1+x^3}\,\mathrm dx.$$ $\endgroup$ – 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.