Math Problem Statement

Determine if the series \( \sum_{n=0}^{\infty} \frac{2n + 5}{(n+2)^2 (n+3)^2} \) converges or diverges and, if it converges, find the sum.

Solution

The problem given is:

n=02n+5(n+2)2(n+3)2\sum_{n=0}^{\infty} \frac{2n + 5}{(n+2)^2 (n+3)^2}

We need to determine whether this series converges or diverges and, if it converges, find its sum.

Step 1: Analyze Convergence

To analyze the convergence of this series, we can consider applying the comparison test or limit comparison test with a simpler series that resembles the given one. Specifically, since the denominator grows as n4n^4 (due to (n+2)2(n+3)2(n+2)^2(n+3)^2), we might want to compare this with a known convergent series that has a similar structure.

Step 2: Simplify the Growth of Terms

For large values of nn, the terms can be approximated as:

2n+5(n+2)2(n+3)22nn4=2n3\frac{2n + 5}{(n+2)^2 (n+3)^2} \approx \frac{2n}{n^4} = \frac{2}{n^3}

So, we can compare this series to 1n3\sum \frac{1}{n^3}, which is a convergent pp-series with p=3>1p = 3 > 1. This suggests that our original series might also converge.

Step 3: Applying the Limit Comparison Test

To confirm convergence, let's apply the limit comparison test with the series 1n3\sum \frac{1}{n^3}:

limn2n+5(n+2)2(n+3)21n3=limn(2n+5)n3(n+2)2(n+3)2\lim_{n \to \infty} \frac{\frac{2n + 5}{(n+2)^2 (n+3)^2}}{\frac{1}{n^3}} = \lim_{n \to \infty} \frac{(2n + 5) n^3}{(n+2)^2 (n+3)^2}

Expanding the denominator for large nn:

(n+2)2n2,(n+3)2n2(n+2)2(n+3)2n4(n+2)^2 \approx n^2, \quad (n+3)^2 \approx n^2 \Rightarrow (n+2)^2 (n+3)^2 \approx n^4

Thus, the expression becomes:

limn(2n+5)n3n4=limn2n+5n=limn2+5n=2\lim_{n \to \infty} \frac{(2n + 5) n^3}{n^4} = \lim_{n \to \infty} \frac{2n + 5}{n} = \lim_{n \to \infty} 2 + \frac{5}{n} = 2

Since this limit is a finite non-zero constant, the limit comparison test tells us that our series behaves similarly to 1n3\sum \frac{1}{n^3}, which is convergent. Therefore, the original series converges.

Step 4: Finding the Sum

The series converges, but finding an exact sum is more complex. This series does not easily simplify to a known form with a closed sum. Therefore, we conclude that the series converges, but without a specific closed-form sum available from standard series techniques.

Final Answer

  • Convergence: Yes, the series converges.
  • Sum: Not easily obtainable in a closed form.

Would you like to see further details, or do you have any additional questions?


Here are some related questions to explore:

  1. How does the pp-series test determine convergence?
  2. What other convergence tests could we apply to this series?
  3. Why does the limit comparison test work in determining convergence?
  4. How would the behavior of the series change if the exponent in the denominator were different?
  5. Is there a way to approximate the sum of the series using numerical methods?

Tip: When unsure about convergence, try to approximate the behavior of the series for large nn and compare it to known series types, such as pp-series or geometric series.

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Math Problem Analysis

Mathematical Concepts

Series
Convergence Tests
Limit Comparison Test

Formulas

\( \sum \frac{1}{n^p} \) for \( p > 1 \)
Limit Comparison Test formula: \( \lim_{n \to \infty} \frac{a_n}{b_n} \)

Theorems

Limit Comparison Test
P-Series Test

Suitable Grade Level

College Level