Evaluating Limit approaching Infinity.

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. 13+23+33+n3n4\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 n3=n2(n+1)24∑n^{3} = \frac{n^{2}(n+1)^{2}}{4}

we get the value of limit as 14\frac{1}{4}

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

13+23+33+n3n4\frac{1^{3} + 2^{3} + 3^{3} + … n^{3}}{n^{4}}

= 13n4+23n4++n3n4\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 14\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

Note by Soham Zemse
5 years, 7 months ago

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You 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 - 5 years, 7 months ago

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Cool explanation! The problem can be also evaluated using Riemann Sums which would be integration from x=0x=0 to x=1x=1, the function x3x^{3}.

Piyal De - 5 years, 7 months ago

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Exactly.

Finn Hulse - 5 years, 7 months ago

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