Past sunday i gave a mathematics olympiad exam at IIT Bombay. I was asked a problem in which i had to find limit of problem :

limit x approaches infinity. \[\frac{1^{3} + 2^{3} + 3^{3} + … n^{3}}{n^{4}}\]

there are 2 ways of evaluating this limit.

1) by using the fact that \[∑n^{3} = \frac{n^{2}(n+1)^{2}}{4}\]

we get the value of limit as \[\frac{1}{4}\]

2) by separating the denominators and using the fact that limit of a sum is the sum of the limits.

\[\frac{1^{3} + 2^{3} + 3^{3} + … n^{3}}{n^{4}}\]

= \[\frac{1^{3}}{n^{4}} + \frac{2^{3}}{n^{4}} + … + \frac{n^{3}}{n^{4}}\]

And it turns out that the value of limit is 0.

I was confused at this while writing the exam. So I choosed \[\frac{1}{4}\] by random.

And I am wondering if I did right decision or not. So I asked here. Please consider explaining me why we get two distinct answers and if my first or second approaches have any errors then please tell me. Thanks

## Comments

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TopNewestYou chose correct answer. The 2nd method is wrong. It is expressed as sum of infinite infinitesimal quantities. It is something like saying 0 multiplied by infinity. – Reaber John · 3 years, 3 months ago

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– Piyal De · 3 years, 3 months ago

Cool explanation! The problem can be also evaluated using Riemann Sums which would be integration from \(x=0\) to \(x=1\), the function \(x^{3}\).Log in to reply

– Finn Hulse · 3 years, 3 months ago

Exactly.Log in to reply