Mwanza W. answered • 04/07/25
Mathematics Expert
To apply the Direct Comparison Test to the given series:
∑n=1∞n+1n2+3,\sum_{n=1}^{\infty} \frac{\sqrt{n}+1}{\sqrt{n^2 + 3}},n=1∑∞n2+3n+1,we need to compare this series to a simpler series whose convergence or divergence we already know. First, let's analyze the behavior of the given term for large nnn.
Step 1: Simplify the terms for large nnn
For large values of nnn, we focus on the dominant terms in both the numerator and the denominator:
- The numerator is n+1\sqrt{n} + 1n+1. For large nnn, n\sqrt{n}n dominates, so n+1∼n\sqrt{n} + 1 \sim \sqrt{n}n+1∼n.
- The denominator is n2+3\sqrt{n^2 + 3}n2+3. For large nnn, n2n^2n2 dominates, so n2+3∼n2=n\sqrt{n^2 + 3} \sim \sqrt{n^2} = nn2+3∼n2=n.
Thus, for large nnn, the general term behaves like:
n+1n2+3∼nn=1n.\frac{\sqrt{n}+1}{\sqrt{n^2 + 3}} \sim \frac{\sqrt{n}}{n} = \frac{1}{\sqrt{n}}.n2+3n+1∼nn=n1.Step 2: Compare with a known series
We compare the given series with the series:
∑n=1∞1n.\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}.n=1∑∞n1.This is a p-series with p=12p = \frac{1}{2}p=21, and we know that a p-series converges if p>1p > 1p>1 and diverges if p≤1p \leq 1p≤1. Since p=12≤1p = \frac{1}{2} \leq 1p=21≤1, this series diverges.
Step 3: Apply the Direct Comparison Test
We now use the Direct Comparison Test. For large nnn, we found that:
n+1n2+3∼1n.\frac{\sqrt{n}+1}{\sqrt{n^2 + 3}} \sim \frac{1}{\sqrt{n}}.n2+3n+1∼n1.Therefore, for large nnn, the given series behaves similarly to a divergent p-series ∑1n\sum \frac{1}{\sqrt{n}}∑n1. Specifically, we can say that for sufficiently large nnn, there exists a constant CCC such that:
1n≤n+1n2+3≤C⋅1n.\frac{1}{\sqrt{n}} \leq \frac{\sqrt{n}+1}{\sqrt{n^2 + 3}} \leq C \cdot \frac{1}{\sqrt{n}}.n1≤n2+3n+1≤C⋅n1.Since the series ∑1n\sum \frac{1}{\sqrt{n}}∑n1 diverges, the Direct Comparison Test tells us that the given series:
∑n=1∞n+1n2+3\sum_{n=1}^{\infty} \frac{\sqrt{n}+1}{\sqrt{n^2 + 3}}n=1∑∞n2+3n+1
diverges.
