Question-135540

Question Number 135540 by mohammad17 last updated on 13/Mar/21

Answered by Olaf last updated on 13/Mar/21

Q 1 a)

1 / S_{n} = \underset{k = 1}{\sum^{n}} [\frac{1}{5^{k}} - \frac{1}{k (k + 1)}]

S_{n} = \underset{k = 1}{\sum^{n}} [\frac{1}{5^{k}} - (\frac{1}{k} - \frac{1}{k + 1})]

S_{n} = \frac{1}{5} . \frac{1 - {(\frac{1}{5})}^{n}}{1 - \frac{1}{5}} - (1 - \frac{1}{n + 1})

S_{n} = - \frac{3}{4} - \frac{1}{4} {(\frac{1}{5})}^{n} + \frac{1}{n + 1}

Double subscripts: use braces to clarify

2 / S_{n} = \underset{k = 0}{\sum^{n}} \frac{{(- 1)}^{k}}{3^{k + 1}} = \frac{1}{3} \underset{k = 0}{\sum^{n}} {(- \frac{1}{3})}^{k}

S_{n} = \frac{1}{3} \times 1 . \frac{1 - {(- \frac{1}{3})}^{n + 1}}{1 - (- \frac{1}{3})}

S_{n} = \frac{1 - {(- \frac{1}{3})}^{n + 1}}{4}

Double subscripts: use braces to clarify

Q 1 b)

1 / S_{n} = \underset{k = 1}{\sum^{n}} \frac{\sqrt{k + 1} - \sqrt{k}}{\sqrt{k^{2} + k}}

S_{n} = \underset{k = 1}{\sum^{n}} \frac{\sqrt{k + 1} - \sqrt{k}}{\sqrt{k} . \sqrt{k + 1}}

S_{n} = \underset{k = 1}{\sum^{n}} [\frac{1}{\sqrt{k}} - \frac{1}{\sqrt{k + 1}}]

S_{n} = 1 - \frac{1}{\sqrt{n + 1}}

Double subscripts: use braces to clarify

2 / S_{n} = \underset{k = 1}{\sum^{n}} [\frac{1}{k} - \frac{1}{k + 2}]

S_{n} = 1 + \frac{1}{2} - \frac{1}{n + 1} - \frac{1}{n + 2}

Double subscripts: use braces to clarify

Answered by Olaf last updated on 13/Mar/21

Q 3 a)

1 / \int \sin x^{3} d x = \int \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \frac{{(x^{3})}^{2 n + 1}}{(2 n + 1)!} d x

= \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \frac{x^{6 n + 4}}{(6 n + 4) (2 n + 1)!}

= \frac{x^{4}}{2} \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \frac{x^{6 n}}{(3 n + 2) (2 n + 1)!}

2 / \int \ln (1 + x) d x = \int \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \frac{x^{n + 1}}{n + 1} d x

= \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \frac{x^{n + 2}}{(n + 1) (n + 2)}

= - x \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n} \frac{x^{n}}{n (n + 1)}

Answered by Olaf last updated on 13/Mar/21

Q 2)

1 / e^{x} = \underset{n = 0}{\sum^{\infty}} \frac{x^{n}}{n!} \Rightarrow \underset{n = 1}{\sum^{\infty}} \frac{e^{n}}{n!} = e^{e} - 1

2 / \underset{k = 2}{\sum^{\infty}} \frac{\ln k}{k - 1} ⩾ \underset{k = 2}{\sum^{\infty}} \frac{\ln 2}{k - 1} = \ln 2 \underset{k = 1}{\sum^{\infty}} \frac{1}{k}

But the harmonic serie Σ \frac{1}{k} diverges

\Rightarrow the serie \underset{k = 2}{\sum^{\infty}} \frac{\ln k}{k - 1} diverges too

3 / \frac{1}{1 - α} = \underset{n = 0}{\sum^{\infty}} α^{n} converges if ∣ α ∣< 1

4 / \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2} + 1} ⩽ \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}}

But the serie Σ \frac{1}{k^{2}} converges (\to \frac{π^{2}}{6})

\Rightarrow the serie \underset{k = 2}{\sum^{\infty}} \frac{\ln k}{k - 1} converges too

5 / 4 \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{4}} converges

6 / \underset{x = 1}{\sum^{\infty}} \frac{\sin^{2} x}{x} \underset{x = 1}{\sum^{\infty}} \frac{1 - \cos (2 x)}{2 x}

But the serie Σ \frac{\cos (2 n)}{n^{2}} converges (Abel)

and the harmonic serie Σ \frac{1}{2 n} diverges

\Rightarrow the serie \underset{x = 1}{\sum^{\infty}} \frac{\sin^{2} x}{x} diverges

Answered by mathmax by abdo last updated on 14/Mar/21

2) \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{3^{n + 1}} = \frac{1}{3} \sum_{n = 0}^{\infty} {(- \frac{1}{3})}^{n} = \frac{1}{3} \times \frac{1}{1 + \frac{1}{3}} = \frac{1}{3} . \frac{3}{4} = \frac{1}{4}