Math Problem Statement

determine convergence of series ((k/(k+1))k)((-1)(k+1))

Solution

To determine the convergence of the series

k=1(kk+1)k(1)k+1,\sum_{k=1}^{\infty} \left( \frac{k}{k+1} \right)^k (-1)^{k+1},

we will apply convergence tests for series. Let's first break it down into components.

1. Form of the series:

The series can be written as:

S=k=1akwhereak=(kk+1)k(1)k+1.S = \sum_{k=1}^{\infty} a_k \quad \text{where} \quad a_k = \left( \frac{k}{k+1} \right)^k (-1)^{k+1}.

Notice that this series has an alternating sign due to the factor (1)k+1(-1)^{k+1}, which suggests that we might use the Alternating Series Test (Leibniz criterion).

2. Alternating Series Test (Leibniz Criterion):

For the series to converge by the alternating series test, two conditions must be satisfied:

  • bk=(kk+1)kb_k = \left( \frac{k}{k+1} \right)^k must be decreasing.
  • limkbk=0\lim_{k \to \infty} b_k = 0.

We will now check these two conditions.

3. Condition 1: Check if bkb_k is decreasing.

We need to check if the sequence bk=(kk+1)kb_k = \left( \frac{k}{k+1} \right)^k is decreasing. To do this, we check the behavior of the terms as kk \to \infty.

First, observe that:

kk+1=11k+1.\frac{k}{k+1} = 1 - \frac{1}{k+1}.

So,

bk=(11k+1)k.b_k = \left( 1 - \frac{1}{k+1} \right)^k.

For large kk, we can approximate this expression using the limit:

limk(11k+1)k=1e.\lim_{k \to \infty} \left( 1 - \frac{1}{k+1} \right)^k = \frac{1}{e}.

Since (11k+1)k\left( 1 - \frac{1}{k+1} \right)^k approaches 1e\frac{1}{e} and does not approach zero initially, it suggests that bkb_k might be decreasing after a certain point.

4. Condition 2: limkbk=0\lim_{k \to \infty} b_k = 0:

As we just observed, the limit of bk=(11k+1)kb_k = \left( 1 - \frac{1}{k+1} \right)^k approaches 1e\frac{1}{e}, which is a positive constant. Therefore,

limkbk=1e0.\lim_{k \to \infty} b_k = \frac{1}{e} \neq 0.

5. Conclusion:

Since the sequence bkb_k does not approach zero, the conditions for the Alternating Series Test are not satisfied.

Therefore, the series does not converge.

Additional Tests

To further confirm, you could apply the Ratio Test or Root Test to analyze whether the series converges absolutely, but since bkb_k does not approach zero, we can already conclude that the series does not converge.

Would you like to explore any other tests or details about this series? Here's a tip:

Tip: If you suspect a series has alternating terms, always check the limit of the non-alternating part first (like bkb_k) to see if it goes to zero, which is a basic requirement for most tests.

Related Questions:

  1. How do you apply the Ratio Test to determine the convergence of a series?
  2. What is the significance of the alternating sign (1)k+1(-1)^{k+1} in the series?
  3. How does the Root Test compare to the Ratio Test for series convergence?
  4. Can you provide an example of a series that converges conditionally but not absolutely?
  5. What is the intuition behind the convergence of series involving terms like (kk+1)k\left( \frac{k}{k+1} \right)^k?

Let me know if you'd like to dive deeper into any of these!

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Math Problem Analysis

Mathematical Concepts

Series Convergence
Alternating Series
Limit of a Sequence

Formulas

Alternating Series Test: If a_n is decreasing and lim (n→∞) a_n = 0, then the series ∑ (-1)^n a_n converges.
Limit Approximation: (1 - 1/(k+1))^k → 1/e as k → ∞

Theorems

Alternating Series Test (Leibniz Criterion)
Limit Properties of Sequences

Suitable Grade Level

Undergraduate Mathematics