Determine whether the following series converge or diverge. If the series converges, then determine its sum.
\sum_{k=0}^{\infty}\left(-\dfrac{3}{4}\right)^k
A geometric series \sum_{k=0}^{\infty}r^k :
- Converges if |r| \lt 1 and its sum is \dfrac{1}{1-r}.
- Diverges if |r|\geq1.
Here we have:
|\dfrac{-3}{4}| \lt 1
Therefore:
\sum_{k=0}^{n}\left(-\dfrac{3}{4}\right)^k converges and the sum equals:
\dfrac{1}{1-\left(\dfrac{-3}{4}\right)}=\dfrac{1}{\dfrac{7}{4}}=\dfrac{4}{7}
\sum_{k=0}^{\infty}\left(-\dfrac{3}{4}\right)^k converges and its sum is \dfrac{4}{7}.
\sum_{k=0}^{\infty}\left(\dfrac{1}{2}\right)^k
A geometric series \sum_{k=0}^{\infty}r^k :
- Converges if |r| \lt 1 and its sum is \dfrac{1}{1-r}.
- Diverges if |r|\geq1.
Here we have:
|\dfrac{1}{2}| \lt 1
Therefore:
\sum_{k=0}^{\infty}\left(\dfrac{1}{2}\right)^k converges and the sum equals:
\dfrac{1}{1-\left(\dfrac{1}{2}\right)}=\dfrac{1}{\dfrac{1}{2}}=2
\sum_{k=0}^{\infty}\left(\dfrac{1}{2}\right)^k converges and the sum is 2.
\sum_{k=0}^{\infty}\left(\dfrac{3}{4}\right)^k
A geometric series \sum_{k=0}^{\infty}r^k :
- Converges if |r| \lt 1 and its sum is \dfrac{1}{1-r}.
- Diverges if |r|\geq1.
Here we have:
|\dfrac{3}{4}| \lt 1
Therefore:
\sum_{k=0}^{n}\left(\dfrac{3}{4}\right)^k converges and the sum equals:
\dfrac{1}{1-\left(\dfrac{3}{4}\right)}=\dfrac{1}{\dfrac{1}{4}}=4
\sum_{k=0}^{n}\left(\dfrac{3}{4}\right)^k converges and its sum equals 4.
\sum_{k=0}^{\infty}\left(-\dfrac{7}{4}\right)^k
A geometric series \sum_{k=0}^{\infty}r^k :
- Converges if |r| \lt 1 and its sum is \dfrac{1}{1-r}.
- Diverges if |r|\geq1.
Here we have:
|\dfrac{-7}{4}| \gt 1
Therefore:
\sum_{k=0}^{\infty}\left(-\dfrac{7}{4}\right)^k diverges.
\sum_{k=0}^{\infty}\left(-1\right)^k
A geometric series \sum_{k=0}^{\infty}r^k :
- Converges if |r| \lt 1 and its sum is \dfrac{1}{1-r}.
- Diverges if |r|\geq1.
Here we have:
|-1| \geq 1
Therefore:
\sum_{k=0}^{\infty}\left(-1\right)^k diverges.
\sum_{k=0}^{\infty}\left(-\dfrac{7}{8}\right)^k
A geometric series \sum_{k=0}^{\infty}r^k :
- Converges if |r| \lt 1 and its sum is \dfrac{1}{1-r}.
- Diverges if |r|\geq1.
Here we have:
|\dfrac{-7}{8}| \lt 1
Therefore:
\sum_{k=0}^{\infty}\left(-\dfrac{7}{8}\right)^k converges and the sum equals:
\dfrac{1}{1-\left(\dfrac{-7}{8}\right)}=\dfrac{1}{\dfrac{15}{16}}=\dfrac{16}{15}
\sum_{k=0}^{\infty}\left(-\dfrac{7}{8}\right)^k converges and its sum equals \dfrac{16}{15}.
\sum_{k=0}^{\infty}\left(\dfrac{1}{4}\right)^k
A geometric series \sum_{k=0}^{\infty}r^k :
- Converges if |r| \lt 1 and its sum is \dfrac{1}{1-r}.
- Diverges if |r|\geq1.
Here we have:
|\dfrac{1}{4}| \lt 1
Therefore:
\sum_{k=0}^{\infty}\left(\dfrac{1}{4}\right)^k converges and the sum equals:
\dfrac{1}{1-\left(\dfrac{1}{4}\right)}=\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}
\sum_{k=0}^{\infty}\left(\dfrac{1}{4}\right)^k converges and its sum is \dfrac{4}{3}.