Задачу просят решить через признаки сравнения. При этом, если пробовать брать для предельного признака сравнения ряды по типу 1/n^2, то никакого конечного числа получить не удается, хотя согласно ответу, этот ряд должен сходится. Что можно сделать в таком случае?
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@eudaimon how would you find a closed form for the following sequence: \sum_{n=1}^\infty \frac{\ln n}{n^{5/4}}? First, give me a hint, do not provide the whole solution. Remember to use dollar signs as LaTeX delimiters.
To find whether the series \sum_{n=1}^\infty \frac{\ln n}{n^{5/4}} converges and possibly find a closed form, we will use comparison tests, as a closed form for such a series is typically not available.
First, recognize that \ln n grows much slower than any power of n. So for large n, \ln n is insignificant compared to n^{p} for any p > 0. Think about comparing the given series with a simpler p-series or a series that you know converges, where the terms are of the form \frac{1}{n^p} for some p > 1.
Hint: You might want to compare the given series with a p-series where the exponent on n is slightly less than 5/4, to use the Limit Comparison Test effectively. Can you find such a series that you know converges and then apply the Limit Comparison Test to conclude about the convergence of the given series?
@eudaimon how would you find a closed form for the following sequence: \sum_{n=1}^\infty \frac{\ln n}{n^{5/4}}? First, give me a hint, do not provide the whole solution. Remember to use dollar signs as LaTeX delimiters.
To find a closed form for the series \sum_{n=1}^\infty \frac{\ln n}{n^{5/4}}, it is unlikely that a closed form exists. Instead, we can focus on determining the convergence of the series using a suitable convergence test.
Hint: Consider using the Integral Test to determine the convergence of the given series. Can you think of a suitable function to integrate and compare it with the series in question?
@eudaimon how would you find a closed form for the following sequence: \sum_{n=1}^\infty \frac{\ln n}{n^{5/4}}? You could test for convergence using comparison tests. Show each step of your reasoning. Remember to use dollar signs as LaTeX delimiters.
To find a closed form for the series \sum_{n=1}^\infty \frac{\ln n}{n^{5/4}}, we will first test for convergence using the Limit Comparison Test.
Let’s consider a simpler series \sum_{n=1}^\infty \frac{1}{n^{5/4}}. We know that this series converges because the exponent 5/4 is greater than 1.
Now, we will take the limit of the ratio between the terms of the given series and the terms of the simpler series:
\lim_{{n \to \infty}} \frac{\frac{\ln n}{n^{5/4}}}{\frac{1}{n^{5/4}}}
Simplifying this expression, we get:
\lim_{{n \to \infty}} \ln n = \infty
Since the limit is positive infinity, this means that the given series and the simpler series have the same convergence behavior.
Therefore, by the Limit Comparison Test, we can conclude that the series \sum_{n=1}^\infty \frac{\ln n}{n^{5/4}} also converges.
However, finding a closed form for this series is unlikely as it involves the logarithm of n.
@eudaimon how would you find a closed form for the following sequence: \sum_{n=1}^\infty \frac{\ln n}{n^{5/4}}? You could test for convergence using comparison tests. Show each step of your reasoning. Remember to use dollar signs as LaTeX delimiters.
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To determine the convergence of the series \sum_{n=1}^\infty \frac{\ln n}{n^{5/4}}, we can employ the Comparison Test or the Limit Comparison Test. Given the presence of the logarithm, the Limit Comparison Test might be more suitable. We would compare our series with a known convergent series. Since \frac{1}{n^{5/4}} is a p-series with p = 5/4 > 1, it is convergent.
Let’s apply the Limit Comparison Test. We will compare our series to the simpler series \sum_{n=1}^\infty \frac{1}{n^{5/4}} by taking the limit of the ratio of their terms:
\lim_{n \to \infty} \frac{\frac{\ln n}{n^{5/4}}}{\frac{1}{n^{5/4}}} = \lim_{n \to \infty} \ln n.
However, there seems to be a mistake here. The limit as n goes to infinity of \ln n is not finite; it diverges. This indicates that I’ve misapplied the Limit Comparison Test. Let’s correct this.
Instead, we should evaluate the following limit:
\lim_{n \to \infty} \frac{\frac{\ln n}{n^{5/4}}}{\frac{1}{n^{5/4}}} = \lim_{n \to \infty} \ln n = \infty.
As the comparison series \frac{1}{n^{5/4}} converges and the limit of the ratio of the terms is infinite, the Limit Comparison Test fails to provide a conclusive result in this form.
However, we can still use the Limit Comparison Test correctly by comparing our series to another series that has a similar behavior for large n. We might instead consider comparing our series with \sum_{n=1}^\infty \frac{1}{n^p} for some p > 1, but with a slight adjustment to account for the logarithmic term.
Let’s take p = 1, making the comparison series \sum_{n=1}^\infty \frac{1}{n}, which is a divergent harmonic series. Now, we can apply the Limit Comparison Test as follows:
\lim_{n \to \infty} \frac{\frac{\ln n}{n^{5/4}}}{\frac{1}{n}} = \lim_{n \to \infty} \frac{\ln n}{n^{1/4}}.
To determine the behavior of this limit, we can use L’Hôpital’s rule or other methods to evaluate the indeterminate form. Would you like to try to evaluate this limit?
@eudaimon yes, evaluate the limit
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Let’s evaluate the limit:
\lim_{n \to \infty} \frac{\ln n}{n^{1/4}}
As n \to \infty, both the numerator and the denominator tend to infinity, which is an indeterminate form \frac{\infty}{\infty}. We can apply L’Hôpital’s Rule, which states that for functions f and g tending to infinity, the limit of \frac{f(n)}{g(n)} as n \to \infty can be found by taking the limit of their derivatives:
\lim_{n \to \infty} \frac{\ln n}{n^{1/4}} = \lim_{n \to \infty} \frac{\frac{d}{dn} (\ln n)}{\frac{d}{dn} (n^{1/4})}.
Taking the derivatives of the numerator and the denominator, we get:
\lim_{n \to \infty} \frac{\frac{1}{n}}{\frac{1}{4}n^{-3/4}} = \lim_{n \to \infty} \frac{4}{n^{1/4}} = 0.
Since the limit is 0, and the comparison series \sum_{n=1}^\infty \frac{1}{n} diverges, the Limit Comparison Test tells us that the original series \sum_{n=1}^\infty \frac{\ln n}{n^{5/4}} converges. This is because if the limit of the ratio of the terms of two series is 0, and the series we are comparing to diverges, then the original series must converge.
Please note that the original series does not have a simple closed form, but we have established its convergence.
@eudaimon I’m not sure I understand why you can claim that \sum 1/n^p is convergent for p>1. Could you elaborate on it? If you need to prove it, do it formally. Also, be concise.
Certainly, the convergence of the series \sum \frac{1}{n^p} for p>1 can be proven using the Integral Test, which states that if f(n) is a positive, continuous, and decreasing function for n \geq 1, then the series \sum_{n=1}^\infty f(n) converges if and only if the integral \int_{1}^\infty f(x)dx converges.
Let’s apply the Integral Test to the function f(x) = \frac{1}{x^p}, where p > 1. We need to evaluate the integral
\int_{1}^\infty \frac{1}{x^p}dx.
This is an improper integral, which we solve as follows:
\int_{1}^\infty \frac{1}{x^p}dx = \lim_{b \to \infty} \int_{1}^{b} \frac{1}{x^p}dx = \lim_{b \to \infty} \left[ \frac{-1}{(p-1)x^{p-1}} \right]_{1}^{b}.
Evaluating this, we get
\lim_{b \to \infty} \left( \frac{-1}{(p-1)b^{p-1}} + \frac{1}{p-1} \right) = \frac{1}{p-1} - \lim_{b \to \infty} \frac{1}{(p-1)b^{p-1}}.
Because p > 1, the term \frac{1}{(p-1)b^{p-1}} goes to 0 as b goes to infinity. Thus, the improper integral converges to \frac{1}{p-1}.
Since the integral converges, by the Integral Test, the series \sum \frac{1}{n^p} also converges for p > 1.