Problem 1 — Series Convergence

Determine the convergence of the following series:

\[\sum_{n=1}^{\infty}\frac{1}{n^2+3}\]

Solution

We use the Limit Comparison Test. Choose the comparison series:

\[\sum_{n=1}^{\infty}\frac{1}{n^2}\]

which is a convergent p-series with $p = 2 > 1$.

Step 1: Compute the limit:

\[L = \lim_{n\to\infty} \frac{a_n}{b_n} = \lim_{n\to\infty} \frac{\dfrac{1}{n^2+3}}{\dfrac{1}{n^2}} = \lim_{n\to\infty} \frac{n^2}{n^2+3}\]

Step 2: Simplify:

\[L = \lim_{n\to\infty} \frac{n^2}{n^2+3} = \lim_{n\to\infty} \frac{1}{1 + \frac{3}{n^2}} = 1\]

Step 3: Since $0 < L = 1 < \infty$ and $\sum \frac{1}{n^2}$ converges, by the Limit Comparison Test:

\[\sum_{n=1}^{\infty}\frac{1}{n^2+3} \quad \textbf{converges.}\]