use limit comparison for this series from 2 to infinity: (19n-1)/(17nsqrtn)

We can apply the **limit comparison test** to determine whether the series 

\[
\sum_{n=2}^{\infty} \frac{19n - 1}{17n \sqrt{n}}
\] 

converges or diverges.

---

### Step 1: Simplify the given series 
For large \(n\), the term \(19n - 1\) behaves similarly to \(19n\). So, the given term is asymptotically similar to:

\[
\frac{19n}{17n \sqrt{n}} = \frac{19}{17\sqrt{n}} = \frac{C}{\sqrt{n}}, \quad \text{where } C = \frac{19}{17}.
\]

Thus, we expect that the given series will behave like the **p-series** \( \sum \frac{1}{\sqrt{n}} \), which is \( \sum \frac{1}{n^{1/2}} \). We now compare our given series with the **p-series**.

---

### Step 2: Apply the Limit Comparison Test
The limit comparison test states that for two positive sequences \(a_n\) and \(b_n\), if:

\[
L = \lim_{n \to \infty} \frac{a_n}{b_n}, 
\]

and \(L\) is a finite, positive number, then both series \( \sum a_n \) and \( \sum b_n \) will either both converge or both diverge.

Let:

\[
a_n = \frac{19n - 1}{17n \sqrt{n}}, \quad b_n = \frac{1}{\sqrt{n}}.
\]

Now, compute the limit:

\[
L = \lim_{n \to \infty} \frac{\frac{19n - 1}{17n \sqrt{n}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{19n - 1}{17n} = \lim_{n \to \infty} \frac{19 - \frac{1}{n}}{17}.
\]

As \(n \to \infty\), \(\frac{1}{n} \to 0\), so:

\[
L = \frac{19}{17}.
\]

---

### Step 3: Conclusion
Since \(L = \frac{19}{17}\) is a finite, positive number, both the given series and the **p-series** \( \sum \frac{1}{\sqrt{n}} \) will behave the same way. 

---

### Step 4: Analyze the p-series
The series \( \sum \frac{1}{\sqrt{n}} = \sum \frac{1}{n^{1/2}} \) diverges because \(p = 1/2 < 1\). 

Thus, the given series:

\[
\sum_{n=2}^{\infty} \frac{19n - 1}{17n \sqrt{n}}
\]

also **diverges** by the limit comparison test.

---

Do you want further details or have any questions?  

Here are 5 related questions you might find useful:  
1. How does the p-series test work, and why does it depend on \(p\)?  
2. Can we use the ratio test instead of limit comparison in this case?  
3. What is the intuition behind the limit comparison test?  
4. What happens if the limit \(L\) equals 0 or infinity in the comparison test?  
5. How do we decide what series to compare with for limit comparison?  

**Tip:** When dealing with series containing polynomial terms, always try to simplify the dominant terms for large \(n\) to find a comparable series.

Use limit comparison for this series from 2 to infinity: (19n-1)/(17nsqrtn)

Learn to apply the Limit Comparison Test to analyze the convergence of the series (19n-1)/(17n√n) from n=2 to infinity. This solution covers simplification, asymptotic comparison with p-series, and the use of the Limit Comparison Test to determine divergence.

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