Math  /  Calculus

QuestionVerify using the indicated test that the infinite series converges. (Hint: Use partial fractions.) n=11n(n+1)\sum_{n=1}^{\infty} \frac{1}{n(n+1)}
By the telescoping series test, we have n=11n(n+1)=limn\sum_{n=1}^{\infty} \frac{1}{n(n+1)}=\lim _{n \rightarrow \infty} \square ) == \square , and thus the series converges.

Studdy Solution

STEP 1

1. We need to verify the convergence of the given infinite series.
2. The series can be expressed in a form that allows the use of the telescoping series test.
3. Partial fraction decomposition is a suitable method to simplify the terms of the series.

STEP 2

1. Decompose the term 1n(n+1)\frac{1}{n(n+1)} using partial fractions.
2. Express the series in a telescoping form.
3. Evaluate the limit of the partial sums as nn approaches infinity to determine convergence.

STEP 3

Decompose the fraction 1n(n+1)\frac{1}{n(n+1)} into partial fractions. Assume:
1n(n+1)=An+Bn+1 \frac{1}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1}
Multiply through by the common denominator n(n+1)n(n+1) to clear the fractions:
1=A(n+1)+Bn 1 = A(n+1) + Bn
Simplify and solve for AA and BB:
1=An+A+Bn 1 = An + A + Bn 1=(A+B)n+A 1 = (A + B)n + A
Equating coefficients, we get:
A+B=0 A + B = 0 A=1 A = 1
From A=1A = 1, substitute into A+B=0A + B = 0:
1+B=0 1 + B = 0 B=1 B = -1
Thus, the partial fraction decomposition is:
1n(n+1)=1n1n+1 \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}

STEP 4

Express the series using the partial fraction decomposition:
n=1(1n1n+1) \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+1} \right)
This series is telescoping, where most terms cancel out.

STEP 5

Write out the first few terms to observe the cancellation:
(1112)+(1213)+(1314)+ \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \cdots
The series simplifies to:
11n+1 1 - \frac{1}{n+1}

STEP 6

Evaluate the limit of the partial sums as nn approaches infinity:
limn(11n+1)=10=1 \lim_{n \to \infty} \left( 1 - \frac{1}{n+1} \right) = 1 - 0 = 1
Since the limit exists and is finite, the series converges.
The series converges to:
1 \boxed{1}

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