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# Super hard calculus question.......

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What is $$\displaystyle\lim_{n\rightarrow \infty}\sum^{n-1}_{k=1}\dfrac{n}{4n^2+k^2}$$

And

What is $$\displaystyle\lim_{n\rightarrow \infty}\sum^{n-1}_{k=1}\dfrac{1}{\sqrt{9n^2-k^2}}$$

MaxWong  Mar 5, 2017

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I guess it's just related to the fact that the integral is defined in terms of the limit of a sum of infinitely small segments.

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Alan  Mar 5, 2017
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Hi Max,

I am not very good at limits so this is probably not correct but this is my best 'guess' :)

I have no idea why this is a calculus question...

I am very happy for you to comment :)

$$\displaystyle\lim_{n\rightarrow \infty}\sum^{n-1}_{k=1}\dfrac{n}{4n^2+k^2}\\~\\ =\displaystyle\lim_{n\rightarrow \infty}\left[ \dfrac{n}{4n^2+1}+\dfrac{n}{4n^2+4}+\dfrac{n}{4n^2+9} \dots \dfrac{n}{4n^2+(n-1)^2}\right]\\ \text{Divide numerator and denominator by }\frac{1}{n^2}\\~\\ =\displaystyle\lim_{n\rightarrow \infty}\left[ \dfrac{(1/n^2)n}{4+1/n^2}+\dfrac{(1/n^2)n}{4+4/n^2}+\dfrac{(1/n^2)n}{4+9/n^2} \dots \dfrac{(1/n^2)n}{4+(n-1)^2/n^2}\right]\\ =\dfrac{0}{4}+\dfrac{0}{4}+\dfrac{0}{4} \dots \dfrac{0}{5}\\ =0$$

The numerator stays zero and the denominator starts at 4 and creeps up to 5 but overall each fraction looks like 0 to me.

Melody  Mar 5, 2017
#2
+26273
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Related to calculus as follows:

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Alan  Mar 5, 2017
#3
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The other one:

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Alan  Mar 5, 2017
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But Alan, integral is sum of all values but sigma is only sum of integer values so how can you go from one to the other like that?

That is not worded properly but I am sure you know what I mean ://

Melody  Mar 5, 2017
#5
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I've never looked into the theory behind it Melody!  Perhaps I should do so.

Alan  Mar 5, 2017
#6
+26273
+6

I guess it's just related to the fact that the integral is defined in terms of the limit of a sum of infinitely small segments.

.

Alan  Mar 5, 2017
#7
+90633
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Thanks Alan,

I shall classify it as one of life's mysteries :))

Melody  Mar 5, 2017

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