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: 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: 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
$$\frac{3n}{\sqrt{n^2 + 3n}} < \sum_{k=1}^{3n} \frac{1}{\sqrt{n^2 + k}} \le \frac{3n}{\sqrt{n^2 + 1}}.$$
A: 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:
$$\sum_{k=1}^{3n}{\frac 1 {\sqrt{n^2 + 3n}}}=\frac {3n} {\sqrt{n^2 + 3n}}<\sum_{k=1}^{3n}{\frac 1 {\sqrt{n^2 + k}}}$$
Hope this helps :)
