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

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