Understanding the Convergence of the Series ∑∞n1cos(n)/n
When dealing with series in mathematics, one often encounters series that exhibit interesting convergence properties. The series in question, ∑∞n1cos(n)/n, is a classic example where the convergence requires a more sophisticated test. This article delves into the convergence of this series and employs Dirichlet's test for its analysis. We will also explore the real part of the sum and derive the final result.
Convergence by Dirichlet's Test
Let us analyze the series ∑∞n1cos(n)/n. To determine its convergence, we will employ Dirichlet's test, which states that a series ∑∞n1anbn converges if the sequence {an} is monotonic and bounded, and the sequence {bn} is such that the partial sums SN ∑1bn are bounded.
Applying Dirichlet's Test to ∑∞n1cos(n)/n
In our case, let an 1/n and bn cos(n).
A. Sequence {1/n}
The sequence an 1/n is positive, monotonically decreasing, and tends to 0 as n approaches infinity. This satisfies the first condition of Dirichlet's test.
B. Sequence {cos(n)}
Now, let's consider the partial sums SN ∑1cos(n). According to the Dirichlet test, we need to show that these partial sums are bounded.
Recall the real part of the sum of the geometric series ( sum_{n1}^infty e^{in} ). Using Euler's formula, ( e^{in} cos(n) isin(n) ), we can write:
[sum_{n1}^N e^{in} sum_{n1}^N (cos(n) isin(n)) cos(1) isin(1) cos(2) isin(2) cdots cos(N) isin(N).]The real part of this sum is:
[Releft( sum_{n1}^N e^{in} right) sum_{n1}^N cos(n).]To find the partial sums, consider the sum of the first N terms of the series:
[sum_{n1}^N e^{in} frac{e^{iN} - 1}{e^i - 1}.]Now, we need to take the real part of this expression:
[Releft( frac{e^{iN} - 1}{e^{i} - 1} right) Releft( frac{cos(N) isin(N) - 1}{cos(1) isin(1) - 1} right).]As N approaches infinity, the sequence of partial sums is bounded because it is the real part of a complex fraction that is bounded. Hence, the second condition of Dirichlet's test is satisfied.
Conclusion: Convergence of the Series
Since the sequence {1/n} is monotonically decreasing and bounded, and the partial sums {cos(n)} are bounded, by Dirichlet's test, the series ∑∞n1cos(n)/n converges.
Real Part of the Sum and the Final Result
To determine the exact value of the sum, we need to consider the limit of the sum as N approaches infinity.
The real part of the sum of the geometric series ∑∞n1ein is:
[sum_{n1}^infty e^{in} -1.]The real part of this is -1. However, we need to take the limit as N goes to infinity:
[Releft( sum_{n1}^infty e^{in} right) -Releft( frac{1}{e^i - 1} right).]Simplifying the expression, we get:
[Releft( frac{1}{e^i - 1} right) Releft( frac{1 - e^{-i} 1}{2 - 2cos(1)} right) frac{2 - 2cos(1)}{2(1 - cos(1))} frac{1}{cos(1/2)}.]Therefore, the sum simplifies to:
[sum_{n1}^infty frac{cos(n)}{n} -Releft( frac{1}{e^i - 1} right) -sqrt{2 - 2cos(1)} 2sinleft(frac{1}{2}right).]Conclusion
The series ∑∞n1cos(n)/n converges to 2sin(1/2). This result relies on the application of Dirichlet's test and the properties of complex exponentials. Understanding such convergence properties is crucial in advanced calculus and analysis, particularly in Fourier series and the study of periodic functions.