The limit of the given function is 0.
Sum of a series
Given the following sum function expressed as:
- [tex]\lim_{n \to \infty} \sum^n_{k=1}(\frac{3}{k(k+2)} )[/tex]
The nth term of the series will be expressed as:
- [tex]\lim_{n \to \infty} \(\frac{3}{k(k+2)}[/tex]
This can also be expressed as [tex]\lim_{n \to \infty} \(\frac{3}{k^2+2k}[/tex]
Divide the resulting nth term of the function by k²
[tex]\lim_{n \to \infty} \(\frac{\frac{3}{n^2} }{\frac{n^2}{n^2} +\frac{2n}{n^2} }\\\lim_{n \to \infty} \(\frac{\frac{3}{n^2} }{1 +\frac{2}{n} }\\[/tex]
Substitute the given value of n into the nth term to have:
[tex]=\(\frac{\frac{3}{\infty^2} }{1 +\frac{2}{\infty} }\\=\frac{0}{1+0}\\= 0[/tex]
Hence the limit of the given function is 0.
Learn more on sum of functions here; https://brainly.com/question/17431959
