Answer :
To solve the series given by
[tex]\[ S_n = \sum_{i=1}^n \frac{1}{i(i+1)} \][/tex]
we can simplify the terms inside the summation.
First, let's rewrite each term in the series. Notice that:
[tex]\[ \frac{1}{i(i+1)} \][/tex]
can be decomposed into partial fractions:
[tex]\[ \frac{1}{i(i+1)} = \frac{A}{i} + \frac{B}{i+1} \][/tex]
To find the constants \(A\) and \(B\), we solve the equation:
[tex]\[ \frac{1}{i(i+1)} = \frac{A}{i} + \frac{B}{i+1} \][/tex]
Multiplying both sides by \( i(i+1) \) gives:
[tex]\[ 1 = A(i+1) + Bi \][/tex]
Simplifying, we get:
[tex]\[ 1 = Ai + A + Bi \][/tex]
[tex]\[ 1 = (A + B)i + A \][/tex]
To satisfy this equation for all \(i\), the coefficients of \(i\) and the constant term must match on both sides. Thus, we have:
[tex]\[ A + B = 0 \][/tex]
[tex]\[ A = 1 \][/tex]
Solving these simultaneously, we get \(A = 1\) and \(B = -1\):
[tex]\[ \frac{1}{i(i+1)} = \frac{1}{i} - \frac{1}{i+1} \][/tex]
So, each term in the series can be rewritten as:
[tex]\[ S_n = \sum_{i=1}^n \left( \frac{1}{i} - \frac{1}{i+1} \right) \][/tex]
This series is telescoping, meaning many terms will cancel out:
Writing out the sum explicitly for the first few terms:
[tex]\[ S_n = \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \cdots + \left( \frac{1}{n} - \frac{1}{n+1} \right) \][/tex]
We notice that most terms cancel out, leaving only:
[tex]\[ S_n = 1 - \frac{1}{n+1} \][/tex]
Thus, the nth partial sum of the series is:
[tex]\[ S_n = 1 - \frac{1}{n+1} \][/tex]
Substituting \(n = 10\):
[tex]\[ S_{10} = 1 - \frac{1}{11} \][/tex]
[tex]\[ S_{10} = 1 - 0.0909090909090909 \][/tex]
[tex]\[ S_{10} = 0.9090909090909091 \][/tex]
Therefore, the sum of the series
[tex]\[ 1 + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{n(n+1)} \][/tex]
for [tex]\( n = 10 \)[/tex] is approximately [tex]\( 0.9090909090909091 \)[/tex].
[tex]\[ S_n = \sum_{i=1}^n \frac{1}{i(i+1)} \][/tex]
we can simplify the terms inside the summation.
First, let's rewrite each term in the series. Notice that:
[tex]\[ \frac{1}{i(i+1)} \][/tex]
can be decomposed into partial fractions:
[tex]\[ \frac{1}{i(i+1)} = \frac{A}{i} + \frac{B}{i+1} \][/tex]
To find the constants \(A\) and \(B\), we solve the equation:
[tex]\[ \frac{1}{i(i+1)} = \frac{A}{i} + \frac{B}{i+1} \][/tex]
Multiplying both sides by \( i(i+1) \) gives:
[tex]\[ 1 = A(i+1) + Bi \][/tex]
Simplifying, we get:
[tex]\[ 1 = Ai + A + Bi \][/tex]
[tex]\[ 1 = (A + B)i + A \][/tex]
To satisfy this equation for all \(i\), the coefficients of \(i\) and the constant term must match on both sides. Thus, we have:
[tex]\[ A + B = 0 \][/tex]
[tex]\[ A = 1 \][/tex]
Solving these simultaneously, we get \(A = 1\) and \(B = -1\):
[tex]\[ \frac{1}{i(i+1)} = \frac{1}{i} - \frac{1}{i+1} \][/tex]
So, each term in the series can be rewritten as:
[tex]\[ S_n = \sum_{i=1}^n \left( \frac{1}{i} - \frac{1}{i+1} \right) \][/tex]
This series is telescoping, meaning many terms will cancel out:
Writing out the sum explicitly for the first few terms:
[tex]\[ S_n = \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \cdots + \left( \frac{1}{n} - \frac{1}{n+1} \right) \][/tex]
We notice that most terms cancel out, leaving only:
[tex]\[ S_n = 1 - \frac{1}{n+1} \][/tex]
Thus, the nth partial sum of the series is:
[tex]\[ S_n = 1 - \frac{1}{n+1} \][/tex]
Substituting \(n = 10\):
[tex]\[ S_{10} = 1 - \frac{1}{11} \][/tex]
[tex]\[ S_{10} = 1 - 0.0909090909090909 \][/tex]
[tex]\[ S_{10} = 0.9090909090909091 \][/tex]
Therefore, the sum of the series
[tex]\[ 1 + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{n(n+1)} \][/tex]
for [tex]\( n = 10 \)[/tex] is approximately [tex]\( 0.9090909090909091 \)[/tex].
