limits sigma notation (1 Viewer)

gamja · Apr 6, 2023

can you do this: $\frac{1}{n}\sum _{k=1}^{\:n}\:\lim _{n\to \infty }\left(\frac{1}{1+\frac{k^2}{n^2}}\right)$

then $\frac{1}{n}\cdot\sum_{k=1}^{n}1$

which equals $\frac{1}{n}?\$

I'm generally not sure how to solve lim questions involving sums... please help! Thank you!

*ps also noticed that latex on BOS is much cleaner - shoutout to whoever cleaned that up

tywebb · Apr 6, 2023

Do it as an integral

$\begin{aligned}\lim\limits_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^n\frac{n^2}{n^2+k^2}&=\lim\limits_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^n\frac{n^2}{n^2+k^2}\div\frac{n^2}{n^2}\\&=\lim\limits_{n\rightarrow\infty}\left(\left(\sum_{k=1}^n\frac{1}{1+\left(\frac{k}{n}\right)^2}\right)\frac{1}{n}\right)\\&=\int_0^1\frac{1}{1+x^2}\ \!dx\\&=[\tan^{-1}x]_0^1\\&=\tan^{-1}1-\tan^{-1}0\\&=\frac{\pi}{4}-0\\&=\frac{\pi}{4}\end{aligned}$

gamja · Apr 6, 2023

could you expand on how you got from line 2->3? (sigma notation to integral) - it looks like the sum of the rectangle areas (trapezoidal rule?) but I'm stuck how the k and n changed to become in terms of 'x'.

tywebb · Apr 6, 2023

Just let $\textstyle\frac{k}{n}=x$ and $\textstyle\frac{1}{n}=\delta x$

As $k$ goes from 1 to $n$ , $x$ goes from $\textstyle\frac{1}{n}$ to $\textstyle\frac{n}{n}$ which as $n\rightarrow\infty$ means $x$ goes from 0 to 1.

And of course in the limit $\delta x$ becomes $dx$

tywebb · Apr 6, 2023

I made a picture if it helps.

limits sigma notation (1 Viewer)

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