$\sum_{n=1}^{\infty} a_n$ is a series and {$s_n$} is the sequence of partial sums. If $s_n = \frac{n^2+1}{4n^2-3}$, find $\sum_{n=1}^{\infty} a_n$

Question

Only idea I have is to use $a_n = s_n-s_{n-1}$, which would mean $a_n = \frac{n^2 + 1}{4n^2 - 3}-\frac{(n-1)^2 + 1}{4(n-1)^2 -3}$. But I'm not sure if I'm thinking in the right direction. Or would $a_n =\lim s_n$? Is that only for when I'm finding the sum?

Answer 1

Thanks to a commenter @MartinR, I was able to find $a_n$. Originally I was confused as to what the question was asking to find $a_n$ as a rule or a single value/sum, and Martin's comment helped me realize that we're looking for a single value.

$$\sum_{n=1}^{\infty} a_n = \lim_{n \rightarrow \infty} s_n = \lim_{n \rightarrow \infty} \frac{n^2 + 1}{4n^2 - 3}.$$

We can multiply and divide the quotient by $n^2$ to simplify it $$\lim_{n \rightarrow \infty} \frac{n^2\cdot\frac{n^2 + 1}{n^2}}{n^2\cdot\frac{4n^2 - 3}{n^2}} = \lim_{n \rightarrow \infty} \frac{1 + \frac{1}{n^2}}{4 - \frac{3}{n^2}}.$$

I know that the quotient of a limit is equal to the quotient of the limits. So $$\lim_{n \rightarrow \infty} \frac{1 + \frac{1}{n^2}}{4 - \frac{3}{n^2}} = \frac{\lim_{n \rightarrow \infty} \left(1+\frac{1}{n^2}\right)}{\lim_{n \rightarrow \infty}\left(4 - \frac{3}{n^2}\right)}.$$

I also know that the limit of a sum/difference is the sum of the limits, we can rewrite what we currently have as: $$\frac{\lim_{n \rightarrow \infty} 1+ \lim_{n \rightarrow \infty}\frac{1}{n^2}}{\lim_{n \rightarrow \infty}4 - \lim_{n \rightarrow \infty}\frac{3}{n^2}}.$$

$\lim_{n \rightarrow \infty} 1 = 1$ and $\lim_{n \rightarrow \infty}\frac{1}{n^2} = 0$, $\lim_{n \rightarrow \infty} 4 = 4$ and $\lim_{n \rightarrow \infty}\frac{3}{n^2} = 0$. So $$\frac{\lim_{n \rightarrow \infty} 1+ \lim_{n \rightarrow \infty}\frac{1}{n^2}}{\lim_{n \rightarrow \infty}4 - \lim_{n \rightarrow \infty}\frac{3}{n^2}} = \frac{1 + 0}{4 - 0} = \frac{1}{4}.$$

In conclusion, $\sum_{n=1}^{\infty} a_n = \frac{1}{4}$.