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

#1**0 **

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

#4**0 **

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