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Part 1
Find a formula for the nth partial sum of the series and use it to find the​ series' sum if the series converges.
StartFraction 11 Over 1 times 2 EndFraction plus StartFraction 11 Over 2 times 3 EndFraction plus StartFraction 11 Over 3 times 4 EndFraction plus midline ellipsis plus StartFraction 11 Over n left parenthesis n plus 1 right parenthesis EndFraction plus midline ellipsis
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Part 1
The formula for the nth partial​ sum, s Subscript n​, of the series is 
  
enter your response here.


To solve this problem, let's analyze the series and find the formula for the \( n \)-th partial sum.

The series given is:
\[
\frac{11}{1 \cdot 2} + \frac{11}{2 \cdot 3} + \frac{11}{3 \cdot 4} + \dots + \frac{11}{n(n+1)} + \dots
\]

This can be expressed as:
\[
\sum_{k=1}^{\infty} \frac{11}{k(k+1)}
\]

### Step 1: Simplify the General Term
Each term of the series is:
\[
\frac{11}{k(k+1)}
\]
To simplify, we can use partial fraction decomposition:
\[
\frac{11}{k(k+1)} = \frac{A}{k} + \frac{B}{k+1}
\]
Multiplying through by \( k(k+1) \), we get:
\[
11 = A(k+1) + Bk
\]
Expanding and matching coefficients, we find \( A = 11 \) and \( B = -11 \). Thus,
\[
\frac{11}{k(k+1)} = \frac{11}{k} - \frac{11}{k+1}
\]

### Step 2: Write the Partial Sum
Now, the \( n \)-th partial sum \( S_n \) is:
\[
S_n = \sum_{k=1}^n \left( \frac{11}{k} - \frac{11}{k+1} \right)
\]
This is a telescoping series, where most terms cancel out. Expanding the sum, we have:
\[
S_n = \left( \frac{11}{1} - \frac{11}{2} \right) + \left( \frac{11}{2} - \frac{11}{3} \right) + \dots + \left( \frac{11}{n} - \frac{11}{n+1} \right)
\]
All intermediate terms cancel, leaving:
\[
S_n = 11 \left(1 - \frac{1}{n+1}\right)
\]
or
\[
S_n = 11 \left(\frac{n}{n+1}\right)
\]

### Step 3: Determine if the Series Converges
To find the sum of the infinite series, we take the limit of \( S_n \) as \( n \to \infty \):
\[
\lim_{n \to \infty} S_n = \lim_{n \to \infty} 11 \left(\frac{n}{n+1}\right) = 11 \times 1 = 11
\]

### Answer
The formula for the \( n \)-th partial sum \( S_n \) is:
\[
S_n = 11 \left(\frac{n}{n+1}\right)
\]
And the sum of the series, if it converges, is:
\[
11
\]

Would you like a breakdown of any part of this solution? 

Here are five related questions for further exploration:
1. How does partial fraction decomposition work in simplifying series terms?
2. What is a telescoping series, and why does it help in finding sums?
3. Can you apply similar techniques to series without a product in the denominator?
4. How can you determine if an infinite series converges or diverges?
5. How would the solution change if the constant numerator (11) were different?

**Tip:** For telescoping series, always check if terms cancel in pairs, which can significantly simplify the summation process.

Find a formula for the nth partial sum of the series and use it to find the series' sum if the series converges. 11/1*2 + 11/2*3 + 11/3*4 + ... + 11/n(n+1) + ...

Learn how to derive the nth partial sum and determine the sum of the converging series 11/1*2 + 11/2*3 + 11/3*4 + ... using partial fraction decomposition and telescoping series. This detailed solution covers step-by-step techniques for finding the sum of an infinite series.

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