Infinite series

9000073402

Level:

In the following list identify the expression which equals

- 1.0 \overset{―}{345}

- 1 - \sum_{n = 1}^{\infty} 345 \cdot 10^{- 3 n - 1}

- 1 - \sum_{n = 1}^{\infty} 345 \cdot 10^{- 3 n}

- \sum_{n = 1}^{\infty} (10 + 345 \cdot 10^{- 3 n - 1})

1 - \sum_{n = 1}^{\infty} 345 \cdot 10^{- 3 n}

9000073403

Level:

Find the sum of the following infinite series.

- 5 \cdot 10^{- 1} - 5 \cdot 10^{- 2} - 5 \cdot 10^{- 3} - 5 \cdot 10^{- 4} - \dots

- 0. \overset{―}{5}

- 0.0 \overset{―}{5}

0

0. \overset{―}{5}

9000073404

Level:

Find the sum of the following infinite series.

\sqrt{2} - 2 + \sqrt{8} - 4 + \sqrt{32} - 8 + \dots

The sum does not exist.

\frac{\sqrt{2}}{1 + \sqrt{2}}

\frac{\sqrt{2}}{1 - \sqrt{2}}

\sqrt{2} - 2

9000073405

Level:

Find the sum of the following infinite series.

\sqrt{2} - 1 + \frac{\sqrt{2}}{2} - \frac{1}{2} + \frac{\sqrt{2}}{4} - \frac{1}{4} + \dots

2 \sqrt{2} - 2

\sqrt{2} - 1

2 \sqrt{2} + 2

\infty

9000073408

Level:

Find the values of

x

which ensure that the following infinite series is convergent.

\sum_{n = 1}^{\infty} \log^{n - 1} x

x \in (\frac{1}{10}; 10)

x \in (1; + \infty)

x \in (1; 10)

x \in R^{+}

9000073407

Level:

Find all the values of

x

such that the following infinite series is convergent.

1 + 3 - 2 x + (3 - 2 x)^{2} + (3 - 2 x)^{3} + \dots

x \in (1; 2)

x \in (- \infty; - 1)

x \in (1; + \infty)

x \in R

9000073406

Level:

Find the sum of the following infinite series.

\sum_{n = 1}^{\infty} {(\frac{\sqrt{2} - 1}{\sqrt{2}})}^{n - 1}

\sqrt{2}

\frac{\sqrt{2} + 1}{\sqrt{2}}

\frac{\sqrt{2}}{2}

Series diverges.

9000063401

Level:

Identify the quotient of the geometric series

\sum_{n = 1}^{\infty} \frac{1}{2^{n - 3}}

\frac{1}{2}

2

1

\frac{1}{8}

9000063402

Level:

Identify the quotient of the geometric series

\sum_{n = 1}^{\infty} 3^{2 - n}

\frac{1}{3}

1

\frac{1}{9}

- \frac{1}{9}

9000063406

Level:

Evaluate the following infinite sum.

\sum_{n = 1}^{\infty} {(- \frac{1}{2})}^{n + 2}

- \frac{1}{12}

- \frac{1}{8}

\frac{1}{2}

1

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