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Convergence and Divergence of Series

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Page Number: - 568
1 / (b, e, g, h)

Page Number :- 576

1 / (c, d), 5 / ( a, c, d, f)

Page Number: - 579

1 /( e , f, h), 3

Please explain all steps.

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Page Number: - 568
1 / (b, e, g, h)

We know that 1+x+x2+x3+x4+... is convergent only when -1 <x<1. And 1+x+x2+x3+x4+...=1/(1-x)
Also we know that removing few terms from the series or multiplying them by a constant does not change the nature of the series
Hence the series 8x+8x2+8x3+8x4+...=8(x+x2+x3+x4+...) will be convergent only when -1< x<1.
The sum is given by 8(x+x2+x3+x4+...)=8(-1+1+x+x2+x3+x4+...)=8(-1+1/(1-x))=8x/(1-x)

Here let t=1/(1+x). The given series can be expressed as 1+t+t2+t3+t4+...which will converge only when -1 <t<1 ie.,
...

Solution Summary

Convergence and divergence of series is investigated. The number of page functions are examined.

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