Question

evaluate
$lim_{n \to \infty } \frac{ {1}^{4} + {2}^{4} + {3}^{4} ... {n}^{4} }{ {n}^{5}} - lim_{n \to \infty }\frac{ {1}^{3} + {2}^{3} + .... {n}^{3}}{ {n}^{5}}$
plz help with the math!!
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small kids don't answer this
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Answer 1

And I go, since nn4+n2+1∼1n3nn4+n2+1∼1n3 and we know that ∑n=∞1n3∑n=∞1n3 converges. so ∑n=1∞nn4+n2+1∑n=1∞nn4+n2+1 is convergent as well.

But I find it hard to calculate the sum. can you give me some hints?

Step-by-step explanation:

As n4+n2+1=(n2+1)2−n2=(n2+1−n)(n2+1+n)n4+n2+1=(n2+1)2−n2=(n2+1−n)(n2+1+n)

and (n2+1+n)−(n2+1−n)=2n(n2+1+n)−(n2+1−n)=2n

nn4+n2+1=12(2n(n2+1−n)(n2+1+n))nn4+n2+1=12(2n(n2+1−n)(n2+1+n))

=12((n2+1+n)−(n2+1−n)(n2+1−n)

Answer 2

This is a quite interesting question ,yet simple to solve if u get the logic or trick behind it.

So, First note that « “If Numerator's degree is 1 less than the denominator's degree ,then , we simply put ( 1/ denominator's degree ).” »

And if « “If Numerator's degree is 2 less than the denominator's degree ,then, answer will be 0.” »

$\green{\mathsf{lim_{n \to \infty } \frac{ {1}^{4} + {2}^{4} + {3}^{4} ... {n}^{4} }{ {n}^{5}} - lim_{n \to \infty }\frac{ {1}^{3} + {2}^{3} + .... {n}^{3}}{ {n}^{5}}}}$

So, Here, $\boxed{\red{\Large{\mathsf{{\dfrac{1}{5}} \:- 0 \:=\: {\dfrac{1}{5}}}}}}$

Answer : 1/5