Find a formula for the nth partial sum of each series and use it to find the series’ sum if the series converges. frac{9}{100}+frac{9}{100^2}+frac{9}{100^3}+...+frac{9}{100^n}+...

Jerold 2021-03-05 Answered
Find a formula for the nth partial sum of each series and use it to find the series’ sum if the series converges.
9100+91002+91003+...+9100n+...
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saiyansruleA
Answered 2021-03-06 Author has 110 answers

Given data
The series is: 9100+91002+91003+...+9100n+...
Consider the infinite series and compare with it given series,
b1+b2+b3+...+bn+...
bn=9100n
The nth partial sum of the given series is,
sn=k=1nbk
=k=1n9100k
The first term of the given series is: b=9100
The common ratio is: r=1100
The expression for the sum of series is,
sn=b(1rn)1r
Substitute the values in the above expression,
sn=(9100)(1(1100)n)11100
=10099(91009100(1100)n)
=10099(91009(1100n+1))
Thus the formula for nth partial sum of series is =10099(91009(1100n+1))
The expression for the limit of the nth partial sum of the given series is,
L=limnsn
=limn(10099(91009limn(1100n+1)))
=10099(91009(1100))
=10099(91009(0))
=111
The nth partial sum of the given series is 111, and the sequence of the partial sum converges to limit L=111, so the given series converges.
Thus the sum of the series is 9100+91002+91003+...+9100n+...=111

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Jeffrey Jordon
Answered 2021-12-16 Author has 2313 answers

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