How to evaluate a sum of the form \u2211 k/n^2 as n goes to infinity

Expert intro: This is one of those limits that looks discrete at first, but it behaves like a scaled area. Once you pull out the right factor, the pattern becomes much clearer.

Detailed walkthrough

We want to evaluate $\lim_{n\to\infty}\sum_{k=1}^{n}\frac{k}{n^2}.$

Step 1: Pull out the constant denominator

Because $n^2$ does not depend on $k$, $\sum_{k=1}^{n}\frac{k}{n^2}=\frac{1}{n^2}\sum_{k=1}^{n}k.$

Step 2: Use the formula for the sum of the first $n$ integers

$\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$ so

=\frac{1}{n^2}\cdot\frac{n(n+1)}{2}$$ ### Step 3: Simplify $$=\frac{n+1}{2n}$$ ### Step 4: Take the limit $$\lim_{n\to\infty}\frac{n+1}{2n}=\frac12$$ So the limit is $$\boxed{\frac12}.$$ ### 💡 Pitfall guide A frequent mistake is treating $\sum_{k=1}^n \frac{k}{n^2}$ like $\frac{1}{n^2}\sum k$ and then stopping too early. You still have to simplify the finite sum formula before taking the limit. Another slip is confusing this with a Riemann sum; it is related in spirit, but the algebra here is enough. ### 🔄 Real-world variant If the sum were $$\sum_{k=1}^n \frac{k}{n^3},$$ then the same method gives $$\frac{1}{n^3}\cdot\frac{n(n+1)}{2}=\frac{n+1}{2n^2},$$ which tends to $0$. If the numerator were $k^2$ instead of $k$, you would use $$\sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6}$$ and the limit would change. ### 🔍 Related terms Riemann sum, sequence limit, sum formula