# limits sigma notation (1 Viewer)

#### gamja

##### Active Member

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

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

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

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

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#### tywebb

##### dangerman
Do it as an integral

\bg_white \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}

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#### gamja

##### Active Member
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

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

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

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

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#### tywebb

##### dangerman
I made a picture if it helps.

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