S=\displaystyle\sum\limits_{n=1}^{\infty }~{\frac{2n+5}{(n-3)(n-2)}}

To evaluate the sum \( S = \sum\limits_{n=1}^{\infty } \frac{2n+5}{(n-3)(n-2)} \), we can first decompose the general term using partial fractions.

### Step 1: Partial Fraction Decomposition
We want to express the fraction \(\frac{2n+5}{(n-3)(n-2)}\) as a sum of partial fractions:
\[
\frac{2n+5}{(n-3)(n-2)} = \frac{A}{n-3} + \frac{B}{n-2}
\]
To find \(A\) and \(B\), we'll solve the equation:
\[
2n + 5 = A(n-2) + B(n-3)
\]
Expanding the right-hand side:
\[
2n + 5 = An - 2A + Bn - 3B = (A + B)n - (2A + 3B)
\]
Now, equate the coefficients of \(n\) and the constant term on both sides:
\[
\begin{aligned}
A + B &= 2, \\
-2A - 3B &= 5.
\end{aligned}
\]

### Step 2: Solve for \(A\) and \(B\)
We now solve the system of equations:
1. \( A + B = 2 \)
2. \( -2A - 3B = 5 \)

From the first equation, solve for \(A\):
\[
A = 2 - B
\]
Substitute this into the second equation:
\[
-2(2 - B) - 3B = 5
\]
Simplify:
\[
-4 + 2B - 3B = 5 \quad \Rightarrow \quad -B - 4 = 5 \quad \Rightarrow \quad B = -9
\]
Substitute \(B = -9\) back into \(A + B = 2\):
\[
A - 9 = 2 \quad \Rightarrow \quad A = 11
\]
So, the partial fraction decomposition is:
\[
\frac{2n+5}{(n-3)(n-2)} = \frac{11}{n-3} - \frac{9}{n-2}
\]

### Step 3: Evaluate the Series
Now the series becomes:
\[
S = \sum_{n=1}^{\infty} \left( \frac{11}{n-3} - \frac{9}{n-2} \right)
\]
This is a telescoping series, but before we proceed, notice that for \(n = 1\), \(n = 2\), and \(n = 3\), the terms \( \frac{11}{n-3} \) are undefined (because the denominator is zero). Hence, the series actually starts from \( n = 4 \):
\[
S = \sum_{n=4}^{\infty} \left( \frac{11}{n-3} - \frac{9}{n-2} \right)
\]
Shifting the index by setting \( m = n-3 \) gives:
\[
S = \sum_{m=1}^{\infty} \left( \frac{11}{m} - \frac{9}{m+1} \right)
\]
This is indeed a telescoping series:
\[
S = \left( \frac{11}{1} - \frac{9}{2} \right) + \left( \frac{11}{2} - \frac{9}{3} \right) + \left( \frac{11}{3} - \frac{9}{4} \right) + \cdots
\]
The terms will cancel out, leaving:
\[
S = 11 \left( 1 + \frac{1}{2} + \frac{1}{3} + \cdots \right) - 9 \left( \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots \right)
\]
The sums remaining are the Harmonic series \( H_n \):
\[
S = 11 H_{\infty} - 9 (H_{\infty} - 1)
\]
This simplifies to:
\[
S = 11H_{\infty} - 9H_{\infty} + 9 = 2H_{\infty} + 9
\]
Since \(H_{\infty}\) diverges to infinity, the series \(S\) diverges as well.

### Final Conclusion
The sum \( S \) diverges because the harmonic series \( H_{\infty} \) diverges.

Would you like more details on the harmonic series, or have any questions?

Here are some related questions:
1. What is a telescoping series and how does it simplify sums?
2. How does partial fraction decomposition work in general?
3. What are the properties of the harmonic series?
4. How can we identify a divergent series?
5. What are the differences between convergent and divergent series?

**Tip:** When working with infinite series, always check if the series converges or diverges before trying to find a sum.

Learn how to evaluate the infinite series sum S = sum_{n=1}^infinity (2n+5)/((n-3)(n-2)) using partial fraction decomposition and telescoping series technique. Understand why the series diverges due to the properties of the harmonic series. Suitable for advanced undergraduate students studying calculus and series convergence.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation