Answer to Question #199056 in Calculus for desmond

Question #199056

1) Find the sum of the geometric series 3 + 2 + 4 3 + 8 9 + . . .


(2) Prove that the following series converges, and find its sum 0.6 + 0.06 + 0.006 + 0.0006 + 0.00006 + · ·


1
Expert's answer
2021-06-01T04:56:02-0400

Solution :-

(a) It should be like this "3+2+\\frac{4}{3}+\\frac{8}{9}....."

a=3

r="\\frac{2}{3}"

Sn="a.\\frac{1-r^n}{1-r}"


"S_n=3.\\frac{(1-\\frac{2}{3}^n)}{(1-\\frac{2}{3})}"

put the value of n and get the sum upto n

terms.

As value of

is not given

So we will find sum of in fininite terms

"= \\frac{a}{1-r}\\\\\n={3\\over 1-\\frac{2}{3}}\\\\\n=9 \\ answer"

(b) 0.6 + 0.06 + 0.006 + 0.0006 + 0.00006 + · ·


"S_n=a.\\frac{1-r^n}{1-r}"


"S_n={\\frac{6}{10}(1-\\frac{1}{10}^n)\\over( 1-\\frac{1}{10})}" .....(1)

put the value of n and get the sum upto n

terms.

If we simplify the Sn

We got

"S_n=\\frac{2}{3}(1-\\frac{1}{10^n})"

And as here also n is not given

So we will find sum of infinite terms

"= \\frac{a}{1-r}\\\\\n={\\frac{6}{10}\\over 1- \\frac{1}{10}}\\\\\n=\\frac{6}{9} \\ answer"


The sum will always come less (using equation 1) than 1 , by the the property of convergence we can say the the series convergence

Hence proved.



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