Exploring the Convergence of Infinite Series: Case Study on the Sequence of ln(n)/n^3

In this article, we delve into the convergence of an important infinite series, specifically the sequence of ln(n)/n^3. We will utilize various testing methods, including the Alternating Series Test and Cauchy's Condensation Test, to determine the type of convergence for the series.

Introduction to the Sequence

We consider the following sequence of positive real numbers:

u_n frac{ln n}{2n^3}

Monotonic Decrease and Limit to Zero

First, we observe that the sequence u_n is monotonically decreasing and approaches zero as n tends to infinity:

lim_{ntoinfty} u_n lim_{ntoinfty} frac{ln n}{2n^3} 0

This is because the logarithmic term grows much slower than the cubic term in the denominator.

Alternating Series Test

Next, we apply the Alternating Series Test to the series:

sum_{n1}^{infty} (-1)^n u_n sum_{n1}^{infty} frac{(-1)^n ln(n)}{2n^3}

For the series to converge, the following conditions must be satisfied:

The series terms alternate in sign. The absolute value of the terms decreases to zero as n to infty. lim_{ntoinfty} left|u_nright| lim_{ntoinfty} frac{ln n}{2n^3} 0

Since these conditions are met, we can conclude that the series converges. This type of convergence is known as conditional convergence, as the series will only converge if the terms alternate in sign, but it does not necessarily imply absolute convergence.

Cauchy's Condensation Test

Let us now apply Cauchy's Condensation Test to determine if the series is absolutely convergent. According to the test, a series sum_{n1}^{infty} u_n converges if and only if the series sum_{n1}^{infty} 2^n u_{2^n} converges.

Substituting the given function u_n frac{ln n}{2n^3} into the test, we get:

v_n 2^n u_{2^n} 2^n cdot frac{ln(2^n)}{2(2^n)^3} frac{ln(2^n)}{2^{2n 1}} frac{n ln 2}{2^{2n 1}}

Ratio Test for Absolute Convergence

To determine if sum_{n1}^{infty} v_n converges, we use the Ratio Test:

ell lim_{n to infty} left| frac{v_{n 1}}{v_n} right|

Calculating this limit step-by-step:

ell lim_{n to infty} left| frac{2^{n 1} cdot frac{(n 1) ln 2}{2^{2(n 1) 1}}}{2^n cdot frac{n ln 2}{2^{2n 1}}} right| lim_{n to infty} frac{2^{n 1} (n 1) ln 2}{2^{2n 3}} cdot frac{2^{2n 1}}{2^n n ln 2} lim_{n to infty} frac{2 (n 1)}{4n} frac{1}{2}

Since ell frac{1}{2} , the series sum_{n1}^{infty} v_n converges. By Cauchy's Condensation Test, the original series sum_{n1}^{infty} u_n is absolutely convergent.

Conclusion

Through a detailed analysis using the Alternating Series Test and the Cauchy's Condensation Test, we have determined that the series:

sum_{n1}^{infty} frac{ln n}{2n^3}

is absolutely convergent. This means the series converges for all n and the convergence is not conditionally, but absolutely.

For further help with such mathematical problems or any other academic assistance, consider utilizing professional writing and editing services. Check out WritersABC for reliable academic solutions.