# Intuition on the concept of bounding a sum

For example, in calculating the following limit

$$\lim_{n\to\infty}\sum_{k=1}^{3n}\frac{1}{\sqrt{n^2+k}}$$

I know (because my professor wrote) that one of the methods of solving this limit is via the sandwich rule since we know the lower and upper bound

$$\frac{3n}{\sqrt{n^2+3n}}<\sum_{k=1}^{3n}\frac{1}{\sqrt{n^2+k}}<\frac{3n}{\sqrt{n^2+1}}$$

But I don't understand why it's so, because I would have guessed it would be

$$\frac{1}{\sqrt{n^2+3n}}<\sum_{k=1}^{3n}\frac{1}{\sqrt{n^2+k}}<\frac{1}{\sqrt{n^2+1}}$$ instead

Your guess is right in terms of bounding but you are forgetting about $$\color{blue}{\text{summing}}$$: $$\color{blue}{\sum_{k=1}^{3n}}\frac{\color{red}1}{\sqrt{n^2+3n}}<\sum_{k=1}^{3n}\frac{1}{\sqrt{n^2+k}}<\color{blue}{\sum_{k=1}^{3n}}\frac{\color{red}1}{\sqrt{n^2+1}} \\ \frac{\color{blue}{3n}}{\sqrt{n^2+3n}}<\sum_{k=1}^{3n}\frac{1}{\sqrt{n^2+k}}<\frac{\color{blue}{3n}}{\sqrt{n^2+1}}$$

A single term of the series obeys the inequality $$\frac{1}{\sqrt{n^2 + 3n}} < \frac{1}{\sqrt{n^2 + k}} \le \frac{1}{\sqrt{n^2 + 1}}$$ for any integer $$k \in \{1, 2, \ldots, 3n\}$$, but the sum does not. Instead, you have to sum each of the expressions accordingly: $$\sum_{k=1}^{3n} \frac{1}{\sqrt{n^2 + 3n}} < \sum_{k=1}^{3n} \frac{1}{\sqrt{n^2 + k}} \le \sum_{k=1}^{3n} \frac{1}{\sqrt{n^2 + 1}},$$ and because the leftmost and rightmost sums are independent of $$k$$, we get

$$\frac{3n}{\sqrt{n^2 + 3n}} < \sum_{k=1}^{3n} \frac{1}{\sqrt{n^2 + k}} \le \frac{3n}{\sqrt{n^2 + 1}}.$$

Since $$k$$ is in the denominator, the bigger $$k$$ the smaller the term of the sum. The biggest term of the sum is therefore $$\frac 1 {\sqrt{n^2 + 1}}$$. If you sum up this term $$3n$$ times, you get something that is definitely bigger than your original sum. But since this sum is independent of $$k$$, you can write it as a simple fraction with $$3n$$ in the numerator.

$$\sum_{k=1}^{3n}{\frac 1 {\sqrt{n^2 + k}}}<\sum_{k=1}^{3n}{\frac 1 {\sqrt{n^2 + 1}}}=\frac {3n} {\sqrt{n^2 + 1}}$$

Similarly, the smallest term of your sum will be $$\frac 1 {\sqrt{n^2 + 3n}}$$. Summing that up and simplifying you get:

$$\sum_{k=1}^{3n}{\frac 1 {\sqrt{n^2 + 3n}}}=\frac {3n} {\sqrt{n^2 + 3n}}<\sum_{k=1}^{3n}{\frac 1 {\sqrt{n^2 + k}}}$$