Free Math Worksheets

- Factor the denominator. - Use partial fractions to write the term as a difference. - Expand several terms and identify the cancellations. - Take the limit of the remaining endpoint terms.

1. Factor the denominator and decompose the term: \(\frac{2}{k(k+2)}=\frac1k-\frac{1}{k+2}\). 2. The \(n\)th partial sum is \(S_n=\sum_{k=1}^{n}\left(\frac1k-\frac{1}{k+2}\right)\). Writing out the terms shows that the series telescopes: \(S_n=1+\frac12-\frac{1}{n+1}-\frac{1}{n+2}\). 3. Take the limit: \(\lim_{n\to\infty}S_n=1+\frac12=\frac32\).

\(\frac32\).

- Factor the difference of squares in the denominator. - Decompose the term into a difference of fractions. - Write the first few terms to expose the telescoping pattern. - Evaluate the remaining expression as \(n\to\infty\).

1. Factor the denominator: \(4k^2-1=(2k-1)(2k+1)\). 2. Use partial fractions: \(\frac{1}{4k^2-1}=\frac12\left(\frac{1}{2k-1}-\frac{1}{2k+1}\right)\). 3. The \(n\)th partial sum telescopes: \(S_n=\frac12\left[\left(1-\frac13\right)+\left(\frac13-\frac15\right)+\cdots+\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)\right]\) \(=\frac12\left(1-\frac{1}{2n+1}\right)\). 4. Therefore, \(\lim_{n\to\infty}S_n=\frac12\).

\(\frac12\).

- Recover a sequence term by subtracting consecutive partial sums. - Decompose the rational expression into partial fractions. - Write several terms of the sum and look for cancellation. - Evaluate the remaining term as \(n\to\infty\).

1. For \(n>1\), \(a_n=S_n-S_{n-1}\) \(=(2n^2+n)-[2(n-1)^2+(n-1)]=4n-1\). For \(n=1\), \(a_1=S_1=3\), which also agrees with the formula. 2. Since \(a_n=4n-1\) and \(a_{n+1}=4n+3\), \(b_n=\frac{1}{(4n-1)(4n+3)}=\frac14\left(\frac{1}{4n-1}-\frac{1}{4n+3}\right)\). 3. The partial sums telescope: \(T_n=\frac14\left[\left(\frac13-\frac17\right)+\left(\frac17-\frac1{11}\right)+\cdots+\left(\frac{1}{4n-1}-\frac{1}{4n+3}\right)\right]\) \(=\frac14\left(\frac13-\frac{1}{4n+3}\right)\). 4. Since \(\frac{1}{4n+3}\to0\), \(\lim_{n\to\infty}T_n=\frac14\cdot\frac13=\frac1{12}\).

1) \(a_n=4n-1\). 2) \(T_n=\frac14\left(\frac13-\frac{1}{4n+3}\right)\), and \(\lim_{n\to\infty}T_n=\frac1{12}\).