Hi Grant!
Great question! We can determine whether this series converges or diverges using the Limit Comparison Test.
We're given:
[
\sum_{n=0}^{\infty}\frac{2n+3}{n^2+3n+1}
]
Step 1: Choose a series to compare it with.
Notice that the numerator has a highest power of (n), while the denominator has a highest power of (n^2).
For large values of (n), the expression behaves approximately like:
[
\frac{2n}{n^2}=\frac{2}{n}
]
This suggests comparing our series with the harmonic series:
[
\sum_{n=1}^{\infty}\frac{1}{n}
]
We know the harmonic series diverges.
Step 2: Apply the Limit Comparison Test.
Let:
[
a_n=\frac{2n+3}{n^2+3n+1}
]
and
[
b_n=\frac{1}{n}
]
Now calculate:
[
L=\lim_{n\to\infty}\frac{a_n}{b_n}
]
Substituting:
[
L=\lim_{n\to\infty}\frac{2n+3}{n^2+3n+1}\cdot n
]
[
L=\lim_{n\to\infty}\frac{2n^2+3n}{n^2+3n+1}
]
Divide the numerator and denominator by (n^2):
[
L=\lim_{n\to\infty}\frac{2+\frac{3}{n}}{1+\frac{3}{n}+\frac{1}{n^2}}
]
As (n\to\infty), all terms containing (1/n) approach zero.
Therefore:
[
\boxed{L=2}
]
Step 3: Interpret the result.
The Limit Comparison Test tells us that if:
[
0<L<\infty
]
then both series either converge or diverge together.
Since (L=2), and the harmonic series diverges, our original series must also diverge.
Final Answer:
[
\boxed{\text{The series DIVERGES.}}
]
Helpful tip: When analyzing a rational expression in a series, compare the highest powers of (n) in the numerator and denominator. This often helps you quickly identify an appropriate comparison series.
Also, starting at (n=0) instead of (n=1) does not change convergence because adding or removing a finite number of terms never changes whether an infinite series converges.
Hope this helps!
Michael
