Computing a Riemann sum with a linear denominator

Key concept: This limit is a classic Riemann-sum problem. The denominator can be rewritten so the sum matches the form of an integral on a short interval.

Step by step

Rewrite the sum in a Riemann-sum form

We are asked to calculate

$\lim_{n\to\infty}\sum_{k=1}^{n} \frac{1}{n+\frac{4k}{5}}.$

Factor out $n$ from the denominator:

$n+\frac{4k}{5}=n\left(1+\frac{4k}{5n}\right).$

So the sum becomes

$\sum_{k=1}^{n} \frac{1}{n\left(1+\frac{4k}{5n}\right)}=\frac{1}{n}\sum_{k=1}^{n}\frac{1}{1+\frac{4k}{5n}}.$

This already looks like a Riemann sum with sample points $x_k=\frac{k}{n}$.

Match it to an integral

Write

$\frac{1}{1+\frac{4k}{5n}}=f\left(\frac{k}{n}\right)$

where

$f(x)=\frac{1}{1+\frac45 x}.$

Then

$\frac{1}{n}\sum_{k=1}^{n} f\left(\frac{k}{n}\right)\to \int_0^1 f(x)\,dx.$

Therefore the limit equals

$\int_0^1 \frac{1}{1+\frac45 x}\,dx.$

Evaluate the integral

Compute

$\int_0^1 \frac{1}{1+\frac45 x}\,dx.$

Let $u=1+\frac45 x$, so $du=\frac45 dx$ and $dx=\frac54 du$.

When $x=0$, $u=1$; when $x=1$, $u=\frac95$.

Thus

=\frac54\int_1^{9/5}\frac{1}{u}\,du =\frac54\ln\left(\frac95\right).$$ So the limit is $$\boxed{\frac54\ln\left(\frac95\right)}.$$ ## Key idea to remember The main skill is turning a sum into a Riemann sum by factoring out the $1/n$ term. Once the expression has the form $$\frac{1}{n}\sum f\left(\frac{k}{n}\right),$$ the limit becomes an integral immediately. ### Pitfall alert A frequent mistake is to treat the denominator as if it were $n\left(1+\frac{4k}{5n}\right)$ and then stop without noticing the essential $1/n$ factor outside the sum. Another error is to guess the interval incorrectly. Because the sample points are $k/n$ from 0 to 1, the integral is over $[0,1]$, not over $[0,5/4]$ or another interval. Be careful not to shift the indexing into a left-endpoint or midpoint form unless you intentionally rewrite the sum that way. ### Try different conditions If the sum were changed to $$\sum_{k=1}^{n}\frac{1}{n+\frac{4k}{5n}},$$ the limit would be completely different, because the denominator would no longer scale linearly with $n$ in the same way. In that variant, the term is close to $1/n$ rather than to a Riemann-sum integrand. A small change in where the $n$ appears can change the entire asymptotic behavior, so it is important to inspect the denominator before matching it to an integral. ### Further reading Riemann sum, definite integral, change of variables