Join the growing community of curious minds on IDNLearn.com and get the answers you need. Whether it's a simple query or a complex problem, our experts have the answers you need.
Sagot :
To determine the conditions under which the series
[tex]\[ \sum_{n=1}^{\infty} \frac{1}{n^p(\ln n)^q} \][/tex]
is convergent, we can follow a step-by-step analysis integrating our knowledge of series and convergence tests.
### Step 1: Understanding the Series
The given series is
[tex]\[ \sum_{n=1}^{\infty} \frac{1}{n^p(\ln n)^q}, \][/tex]
which includes both a polynomial term [tex]\( n^p \)[/tex] and a logarithmic term [tex]\( (\ln n)^q \)[/tex] in the denominator.
### Step 2: Review of p-Series
Recall the p-series:
[tex]\[ \sum_{n=1}^{\infty} \frac{1}{n^p}. \][/tex]
This series converges if and only if [tex]\( p > 1 \)[/tex]. This serves as a starting point to understand the convergence behavior of our given series by considering the [tex]\( n^p \)[/tex] term in isolation.
### Step 3: Consideration of the Logarithmic Term
When analyzing the convergence of the given series, the addition of the [tex]\( (\ln n)^q \)[/tex] factor in the denominator affects the rate of convergence. A logarithmic term grows much slower than any polynomial term, which means:
- The term [tex]\( (\ln n)^q \)[/tex] being in the denominator makes the terms of the series smaller.
- If [tex]\( p > 1 \)[/tex], the series [tex]\( \sum \frac{1}{n^p} \)[/tex] converges, and the presence of [tex]\( (\ln n)^q \)[/tex] generally acts to further ensure convergence.
### Step 4: Applying the Comparison Test
To formally argue for the convergence, we can use the Comparison Test:
- If the series [tex]\( \sum \frac{1}{n^p} \)[/tex] converges for [tex]\( p > 1 \)[/tex], and
- [tex]\( \frac{1}{n^p(\ln n)^q} \leq \frac{1}{n^p} \)[/tex] for sufficiently large [tex]\( n \)[/tex],
Then the given series converges because the additional multiplicative factor [tex]\( \frac{1}{(\ln n)^q} \)[/tex] can only decrease the term values further.
### Step 5: Conditions on [tex]\( p \)[/tex] and [tex]\( q \)[/tex]
Given that [tex]\( (\ln n)^q \)[/tex] in the denominator generally supports convergence:
- For [tex]\( n^p \)[/tex] Term: [tex]\( p \)[/tex] must be greater than 1 ([tex]\( p > 1 \)[/tex]) to ensure the p-series [tex]\(\sum_{n=1}^{\infty} \frac{1}{n^p}\)[/tex] converges.
- For [tex]\( (\ln n)^q \)[/tex] Term: [tex]\( q \)[/tex] must be greater than 0 ([tex]\( q > 0 \)[/tex]) to ensure it has a positive effect in reducing the series terms.
### Conclusion
Thus, the given series
[tex]\[ \sum_{n=1}^{\infty} \frac{1}{n^p(\ln n)^q} \][/tex]
converges when [tex]\( p \)[/tex] and [tex]\( q \)[/tex] satisfy the following conditions:
- [tex]\( p \)[/tex] must be greater than 1 ([tex]\( p > 1 \)[/tex]), and
- [tex]\( q \)[/tex] must be greater than 0 ([tex]\( q > 0 \)[/tex]).
So, the series converges if [tex]\( p > 1 \)[/tex] and [tex]\( q > 0 \)[/tex].
[tex]\[ \sum_{n=1}^{\infty} \frac{1}{n^p(\ln n)^q} \][/tex]
is convergent, we can follow a step-by-step analysis integrating our knowledge of series and convergence tests.
### Step 1: Understanding the Series
The given series is
[tex]\[ \sum_{n=1}^{\infty} \frac{1}{n^p(\ln n)^q}, \][/tex]
which includes both a polynomial term [tex]\( n^p \)[/tex] and a logarithmic term [tex]\( (\ln n)^q \)[/tex] in the denominator.
### Step 2: Review of p-Series
Recall the p-series:
[tex]\[ \sum_{n=1}^{\infty} \frac{1}{n^p}. \][/tex]
This series converges if and only if [tex]\( p > 1 \)[/tex]. This serves as a starting point to understand the convergence behavior of our given series by considering the [tex]\( n^p \)[/tex] term in isolation.
### Step 3: Consideration of the Logarithmic Term
When analyzing the convergence of the given series, the addition of the [tex]\( (\ln n)^q \)[/tex] factor in the denominator affects the rate of convergence. A logarithmic term grows much slower than any polynomial term, which means:
- The term [tex]\( (\ln n)^q \)[/tex] being in the denominator makes the terms of the series smaller.
- If [tex]\( p > 1 \)[/tex], the series [tex]\( \sum \frac{1}{n^p} \)[/tex] converges, and the presence of [tex]\( (\ln n)^q \)[/tex] generally acts to further ensure convergence.
### Step 4: Applying the Comparison Test
To formally argue for the convergence, we can use the Comparison Test:
- If the series [tex]\( \sum \frac{1}{n^p} \)[/tex] converges for [tex]\( p > 1 \)[/tex], and
- [tex]\( \frac{1}{n^p(\ln n)^q} \leq \frac{1}{n^p} \)[/tex] for sufficiently large [tex]\( n \)[/tex],
Then the given series converges because the additional multiplicative factor [tex]\( \frac{1}{(\ln n)^q} \)[/tex] can only decrease the term values further.
### Step 5: Conditions on [tex]\( p \)[/tex] and [tex]\( q \)[/tex]
Given that [tex]\( (\ln n)^q \)[/tex] in the denominator generally supports convergence:
- For [tex]\( n^p \)[/tex] Term: [tex]\( p \)[/tex] must be greater than 1 ([tex]\( p > 1 \)[/tex]) to ensure the p-series [tex]\(\sum_{n=1}^{\infty} \frac{1}{n^p}\)[/tex] converges.
- For [tex]\( (\ln n)^q \)[/tex] Term: [tex]\( q \)[/tex] must be greater than 0 ([tex]\( q > 0 \)[/tex]) to ensure it has a positive effect in reducing the series terms.
### Conclusion
Thus, the given series
[tex]\[ \sum_{n=1}^{\infty} \frac{1}{n^p(\ln n)^q} \][/tex]
converges when [tex]\( p \)[/tex] and [tex]\( q \)[/tex] satisfy the following conditions:
- [tex]\( p \)[/tex] must be greater than 1 ([tex]\( p > 1 \)[/tex]), and
- [tex]\( q \)[/tex] must be greater than 0 ([tex]\( q > 0 \)[/tex]).
So, the series converges if [tex]\( p > 1 \)[/tex] and [tex]\( q > 0 \)[/tex].
We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. Your search for answers ends at IDNLearn.com. Thanks for visiting, and we look forward to helping you again soon.
