Question 933251
Each term of <pre>1/3, 1/3^2, 1/3^3, 1/3^4, 1/3^5, ...</pre> is in the form {{{a*(r)^(n-1)}}} where {{{a = 1/3}}} and {{{r = 1/3}}}



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{{{S[n] = a*((1-r^n)/(1-r))}}} Geometric series nth partial sum formula.



{{{S[n] = (1/3)*((1-(1/3)^n)/(1-1/3))}}} Plug in {{{r = 1/3}}}



{{{S[n] = (1/3)*((1-(1/3)^n)/(2/3))}}}



{{{S[n] = (1/3)*(3/2)*(1-(1/3)^n)}}}



{{{S[n] = (1/2)*(1-(1/3)^n)}}}



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Fifth Partial Sum (n = 5)



{{{S[n] = (1/2)*(1-(1/3)^n)}}}



{{{S[5] = (1/2)*(1-(1/3)^5)}}}



{{{S[5] = (1/2)*(1-1/243)}}}



{{{S[5] = (1/2)*(242/243)}}}



{{{S[5] = 121/243}}}



The fifth partial sum is {{{121/243}}} which is approximately {{{0.49794238683128



}}} (I'd stick to the exact answer whenever you can)



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Sixth Partial Sum (n = 6)



{{{S[n] = (1/2)*(1-(1/3)^n)}}}



{{{S[6] = (1/2)*(1-(1/3)^6)}}}



{{{S[6] = (1/2)*(1-1/729)}}}



{{{S[6] = (1/2)*(728/729)}}}



{{{S[6] = 364/729}}}



The sixth partial sum is exactly {{{364/729}}} which is approximately {{{0.49931412894376}}}



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