Question

Determine convergence or divergence. (a) \(\sum \frac{2+\sin n \theta}{n^{2}+\sin n \theta}\) (b) \(\sum \frac{n+1}{n} r^{n}(r>0)\) (c) \(\sum e^{-n \rho} \cosh n \rho(\rho>0)\) (d) \(\sum \frac{n+\log n}{n^{2}(\log n)^{2}}\) (e) \(\sum \frac{n+\log n}{n^{2} \log n}\) (f) \(\sum \frac{(1+1 / n)^{n}}{2^{n}}\)

Short answer

In summary, the convergence or divergence of the given series is as follows: (a) The series \(\sum \frac{2+\sin n \theta}{n^{2}+\sin n \theta}\) converges by the Limit Comparison Test. (b) The series \(\sum \frac{n+1}{n} r^{n}(r>0)\) converges if \(r < 1\) and diverges if \(r \ge 1\) by the Ratio Test. (c) The series \(\sum e^{-n \rho} \cosh n \rho(\rho>0)\) converges by the Comparison Test. (d) The series \(\sum \frac{n+\log n}{n^{2}(\log n)^{2}}\) converges by the Limit Comparison Test. (e) The series \(\sum \frac{n+\log n}{n^{2} \log n}\) diverges by the Comparison Test. (f) The series \(\sum \frac{(1+1 / n)^{n}}{2^{n}}\) converges by the Ratio Test.

Step-by-step solution

Limit Comparison Test

We will compare our given series with a simpler one, \(\sum \frac{1}{n^2}\). Let's call our original series \(a_n\) and the simplified series \(b_n\). We need to compute the limit as n goes to infinity: \(\lim_{n\to\infty}\frac{a_n}{b_n} = \lim_{n\to\infty}\frac{\frac{2+\sin n \theta}{n^{2}+\sin n \theta}}{\frac{1}{n^2}}\) After simplification, we get: \(\lim_{n\to\infty}\frac{2+\sin n \theta}{1+\frac{\sin n \theta}{n^2}}\) Since the sine function is bounded between -1 and 1, the limit converges to 2, as n goes to infinity. Since our simplified series, \(\sum \frac{1}{n^2}\), converges (it is a p-series with p=2), our given series also converges. (b) Convergence or divergence of \(\sum \frac{n+1}{n} r^{n}(r>0)\)

Ratio Test

Let's apply the Ratio Test to this series. We need to compute the limit as n goes to infinity of the ratio of consecutive terms: \(\lim_{n\to\infty}\frac{\frac{n+2}{n+1}r^{n+1}}{\frac{n+1}{n}r^n}\) Simplifying the expression, we get: \(\lim_{n\to\infty}\frac{(n+2)n}{(n+1)^2}r\) The expression in the limit converges to \(r\) as n goes to infinity. Hence, the series converges if \(r < 1\) and diverges if \(r > 1\). If \(r = 1\), the series becomes a harmonic series, \(\sum \frac{n+1}{n}\), which also diverges. (c) Convergence or divergence of \(\sum e^{-n \rho} \cosh n \rho(\rho>0)\)

Comparison Test

We will use the Comparison Test to check the convergence of our series. Also, we know that \(\cosh x \ge 1\), for every value of x. Therefore, we can state that: \(e^{-n \rho} \cosh n \rho \ge e^{-n \rho}\) Since the series \(\sum e^{-n \rho}\) converges (it is a convergent geometric series), by the Comparison Test, our given series also converges. (d) Convergence or divergence of \(\sum \frac{n+\log n}{n^{2}(\log n)^{2}}\)

Limit Comparison Test

Comparing our given series with \(\sum \frac{1}{n(\log n)^2}\), we compute the limit as n goes to infinity: \(\lim_{n\to\infty}\frac{\frac{n+\log n}{n^{2}(\log n)^{2}}}{\frac{1}{n(\log n)^2}}\) Simplifying, we get: \(\lim_{n\to\infty}\frac{n+\log n}{n^2}\) Evaluating the limit yields 0, which means that both series have the same convergence behavior. It is known that \(\sum \frac{1}{n(\log n)^2}\) converges, so our given series also converges. (e) Convergence or divergence of \(\sum \frac{n+\log n}{n^{2} \log n}\)

Comparison Test

Comparing our given series with \(\sum \frac{1}{n \log n}\), using the Comparison Test, we find that our given series converges if the sum \(\sum \frac{1}{n \log n}\) converges. It is known that the series \(\sum \frac{1}{n \log n}\) diverges (by the Integral Test). Therefore, our given series also diverges. (f) Convergence or divergence of \(\sum \frac{(1+1 / n)^{n}}{2^{n}}\)

Ratio Test

Applying the Ratio Test to this series, we compute the limit as n goes to infinity of the ratio of consecutive terms: \(\lim_{n\to\infty}\frac{(1+\frac{1}{n+1})^{n+1}}{(1+\frac{1}{n})^n}\frac{1}{2}\) The limit converges to \(\frac{1}{2e}\), since \((1+\frac{1}{n})^n\) converges to 'e' as n goes to infinity. Since the limit of the Ratio Test is less than 1, our given series converges.

Key concepts

Limit Comparison Test

The Limit Comparison Test is a powerful tool to determine the convergence or divergence of a series by comparing it to another series with known behavior. This method is often used when direct comparison is cumbersome or when the terms of the series have complex expressions.

In practice, you select a benchmark series, typically a well-known series like a p-series, and compute the limit of the ratio of terms from both series:

• If this limit is a positive finite number, both series will exhibit the same convergence behavior.

For example, consider the series \[ ext{original series } a_n = rac{2+ ext{sin} n heta}{n^{2}+ ext{sin} n heta} \] and the simpler p-series \[ b_n = rac{1}{n^2} \].
Compute the limit:\[ ext{lim}_{n \to \infty} \frac{a_n}{b_n} = ext{lim}_{n \to \infty} \frac{2+ ext{sin} n heta}{1 + \frac{ ext{sin} n heta}{n^2}} = 2 \] This calculation shows the original series converges because the benchmark series converges and the limit is a positive finite number.
This test simplifies the analysis by breaking it down into manageable comparisons.

Ratio Test

The Ratio Test is commonly used for series where terms involve factorials or exponentials. It's particularly handy for series where the terms grow or shrink rapidly.

To apply the Ratio Test, analyze the limit of the absolute value of the ratio of consecutive terms:

• If this limit is less than 1, the series converges absolutely.
• If the limit is greater than 1, the series diverges.
• If the limit equals 1, the test is inconclusive.

Consider the series:\[ ext{original series } a_n = rac{n+1}{n} r^{n} \]
Compute the limit:\[ \text{lim}_{n \to \infty} \frac{|a_{n+1}|}{|a_n|} = \text{lim}_{n \to \infty} \frac{(n+2)n}{(n+1)^2} r = r \].

This implies the series:- Converges if \( r < 1 \) - Diverges if \( r > 1 \).

For \( r = 1 \), the Ratio Test does not provide insights, as the series becomes harmonic, which naturally diverges.

Comparison Test

The Comparison Test uses another series with known convergence properties to determine the behavior of a given series. This method can be very intuitive as it relies on the natural order of growth of the terms.

There are two cases when using the Comparison Test:

• If each term of a convergent series is greater than or equal to each term of the series you're testing, then the series converges.
• If each term of a divergent series is less than or equal to each term of the series you're testing, then the series diverges.

For example: Consider\[ ext{original series } a_n = e^{-n \rho} \cosh n \rho \]It can be compared with \[ b_n = e^{-n \rho} \] because we know: \( \cosh n \rho \geq 1 \).
The comparison shows each term of the given series is greater than or equal to the series \( b_n \), which converges as a geometric series with ratio less than 1. Thus, through the Comparison Test, we confirm that the original series converges.

Utilizing these tests effectively uncovers the convergence behavior of complex series by simplifying their analytical path.