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Determine if the series converges or diverges. If it converges, determine its sum.
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(a) \(\displaystyle \sum_{n=0}^{\infty} 6 \left( \frac{2}{3} \right)^n \)
(b) \(\displaystyle \sum_{n=2}^{\infty} 8 \left( -\frac{1}{2} \right)^n \)
(c) \(\displaystyle \sum_{n=1}^{\infty} 8 \left( -\frac{3}{4} \right)^n \)
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(a) \(\displaystyle \sum_{n=2}^{\infty} \frac{5}{2^n} \)
(b) \(\displaystyle \sum_{n=1}^{\infty} \frac{6}{(-3)^n} \)
(c) \(\displaystyle \sum_{n=1}^{\infty} \frac{4}{2^{-n}} \)
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(a) \(\displaystyle \sum_{n=1}^{\infty} \ln \left( \frac{n^2+5}{3n^2+1} \right) \)
(b) \(\displaystyle \sum_{n=1}^{\infty} \left( \frac{1}{5^n} + \frac{5}{n} \right) \)
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(a) \(\displaystyle \sum_{n=1}^{\infty} \frac{7}{\pi^{n}} \)
(b) \(\displaystyle \sum_{n=1}^{\infty} 8^{n+1} 9^{-n} \)
(c) \(\displaystyle \sum_{n=1}^{\infty} \frac{(-3)^{n-1}}{5^{n}} \)
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(a) \(\displaystyle \frac{3}{2} + \frac{3}{4} + \frac{3}{6} + \frac{3}{8} + \cdots \)
(b) \(\displaystyle \frac{1}{32} - \frac{1}{16} + \frac{1}{8} - \frac{1}{4} + \cdots \)
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\(\displaystyle 2 - 1 + \frac{1}{2} - \frac{1}{2} + \frac{1}{3} - \frac{1}{3} + \frac{1}{4} - \frac{1}{4} + \cdots \)
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\(\displaystyle \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+1} \right) \)
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\(\displaystyle \sum_{n=3}^{\infty} \left( \frac{1}{n} - \frac{1}{n+2} \right) \)
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\(\displaystyle \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+2} \right) \)
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\(\displaystyle \sum_{n=1}^{\infty} \left( e^{\frac{1}{n}} - e^{\frac{1}{n+1}} \right) \)
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\(\displaystyle \sum_{n=1}^{\infty} \left( 2^{\frac{1}{n}} - 2^{\frac{1}{n+1}} \right) \)
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\(\displaystyle \sum_{n=1}^{\infty} \left( 3^{n} - 3^{n+1} \right) \)
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\(\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2 + n} \)
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\(\displaystyle \sum_{n=2}^{\infty} \frac{n^2 + 6n + 3}{2n^2 + n - 1} \)
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Write the series as a telescoping sum, find a formula for \( S_N \), and find the sum of the series (if it converges).
\(\displaystyle \sum_{n=2}^{\infty} \ln \left( 1 + \frac{1}{n} \right) \)
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