+9519 What is \(\displaystyle\lim_{n\rightarrow \infty}\sum^{n-1}_{k=1}\dfrac{n}{4n^2+k^2}\)
And
What is \(\displaystyle\lim_{n\rightarrow \infty}\sum^{n-1}_{k=1}\dfrac{1}{\sqrt{9n^2-k^2}}\)
+118608 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.
+118608 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 ://
+33615 I've never looked into the theory behind it Melody! Perhaps I should do so.
