Math Problem Statement

Which of the following series is convergent?

Solution

The image contains a mathematical question about series convergence. The series options are:

(a) n=11n+1n\sum_{n=1}^{\infty} \frac{1}{\sqrt{n+1} - \sqrt{n}}
(b) n=1sinnn2\sum_{n=1}^{\infty} \frac{\sin n}{n^2}
(c) n=1(1)nlogn\sum_{n=1}^{\infty} (-1)^n \log n
(d) n=1lognn\sum_{n=1}^{\infty} \frac{\log n}{n}

To determine which series is convergent, let’s analyze each one:

  1. (a) n=11n+1n\sum_{n=1}^{\infty} \frac{1}{\sqrt{n+1} - \sqrt{n}}
    To simplify, rationalize the denominator: 1n+1nn+1+nn+1+n=n+1+n(n+1)n=n+1+n.\frac{1}{\sqrt{n+1} - \sqrt{n}} \cdot \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}} = \frac{\sqrt{n+1} + \sqrt{n}}{(n+1) - n} = \sqrt{n+1} + \sqrt{n}. This series behaves similarly to n=1n\sum_{n=1}^{\infty} \sqrt{n}, which diverges because n\sqrt{n} grows without bound.

  2. (b) n=1sinnn2\sum_{n=1}^{\infty} \frac{\sin n}{n^2}
    The function sinn\sin n is bounded between -1 and 1, so this series is comparable to n=11n2\sum_{n=1}^{\infty} \frac{1}{n^2}, which converges by the p-series test with p=2>1p = 2 > 1.

  3. (c) n=1(1)nlogn\sum_{n=1}^{\infty} (-1)^n \log n
    This series is an alternating series, but logn\log n grows without bound. Therefore, this series does not satisfy the conditions for the alternating series test and diverges.

  4. (d) n=1lognn\sum_{n=1}^{\infty} \frac{\log n}{n}
    This is similar to the series n=11n\sum_{n=1}^{\infty} \frac{1}{n} (harmonic series), but with a logarithmic term. This series diverges, as logn\log n grows slower than nn, but still causes divergence.

Conclusion:

The only convergent series is (b) n=1sinnn2\sum_{n=1}^{\infty} \frac{\sin n}{n^2}.

Would you like more details on any of these steps, or have any other questions related to series convergence? Here's a tip:

Tip: To test series convergence, try using comparison tests, the integral test, or the ratio test to assess whether a series converges or diverges based on its behavior as nn \to \infty.


Some additional related questions you might find interesting:

  1. How does the p-series test determine convergence for series like 1np\sum \frac{1}{n^p}?
  2. What is the alternating series test, and how can it be applied?
  3. How does the ratio test work for determining convergence?
  4. How do logarithmic terms affect the convergence of series?
  5. What are the differences between absolute and conditional convergence in series?

Let me know if you'd like any further elaboration!

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Math Problem Analysis

Mathematical Concepts

Series Convergence
Rationalization of Denominators
Alternating Series
p-Series Test
Comparison Test

Formulas

Rationalization: a / (b - c) = a(b + c) / (b^2 - c^2)
p-Series Test: ∑ 1/n^p converges if p > 1
Alternating Series Test
Divergence Criterion

Theorems

p-Series Test
Comparison Test
Alternating Series Test

Suitable Grade Level

Undergraduate Mathematics or Advanced High School