To determine whether the series

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

is convergent, we use comparison tests and analyze the general term.

Step 1: Compare with a known convergent series

The term $\frac{3}{n^2 + 3}$ is less than $\frac{3}{n^2}$ for all $n \geq 1$. The series $\sum \frac{3}{n^2}$ is simply a constant multiple of $\sum \frac{1}{n^2}$, which is a p-series with $p = 2 > 1$.

Since p-series converge for $p > 1$, the series $\sum_{n=1}^\infty \frac{3}{n^2}$ converges.

Step 2: Apply the comparison test

• Since $\frac{3}{n^2 + 3} \leq \frac{3}{n^2}$ for all $n$,
• And since $\sum \frac{3}{n^2}$ converges,
• By the comparison test, the series $\sum_{n=1}^\infty \frac{3}{n^2 + 3}$ also converges.

Conclusion:

The series $\sum_{n=1}^\infty \frac{3}{n^2 + 3}$ is convergent.

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Would you like further explanation of any step or related examples?

Related Questions:

1. How do p-series behave for different values of $p$?
2. What is the comparison test, and how does it work?
3. Can you find the sum of a convergent series like this?
4. How do we determine divergence or convergence using the ratio test?
5. What other tests can we use to confirm convergence?

Tip: When comparing series, always choose a simpler one whose behavior you know well!