Series, Calculating the limit , if does not exist prove it.

Question

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

I'm having a bit of difficulty trying to prove why the limit of this expression is infinity (well that what is my intuition says), may i have your help?

Answer 1

Can you see why this expression approximates $n$? Hint: each fraction approximates $1$ for sufficiently large numerators.

Answer 2

$$\frac{k}{k+3} \ge \frac12 \iff 2k \ge k+3 \iff k \ge 3$$

$$\frac{n}{n+2} \ge \frac12 \iff 2n \ge n+2 \iff n \ge 2$$

If $n \ge 3$, $$\frac{n}{n+2}\sum_{k=1}^n\frac{k}{k+3} \ge \frac{n}{n+2}\sum_{k=3}^n\frac{k}{k+3} \ge \frac12 \sum_{k=3}^n\frac12 = \frac{n-2}{4} \xrightarrow[n \to \infty]{} \infty.$$