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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].
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