Determine if the series is convergent?
summation of n/(10n+3), n-->infinite
Solution
To check if the series converges or diverges, we\'ll have to evaluate the limit of the series.
Let an = Sum n/(10n + 3)
lim an = lim n/(10n + 3)
lim an = lim n/n*(10 + 3/n)
We\'ll simplify and we\'ll get:
lim an = lim 1/(10 + 3/n)
Since the limit of the fraction 3/n converges to zero, we\'ll have:
lim an = 1/10
Since the value of the limit is 1/10, therefore the series converges to 1/10..
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Determine the AW of costs and the ESL for a machine that has a first.pdf
1. Determine the AW of costs and the ESL for a machine that has a first cost of $11500 at an
interest rate of 10.4% given the operating costs and year-end salvage values shown below.
ESL = yearsYearAOC, $Salvage, $AW of Costs1-205077052-205052393-205039294-
205029475-20501974
Solution
Year AOC, $ Salvage, $ AW of Costs 1 -2050 7705 -6377.72 2 -2050 5239 -5554.81 3 -
2050 3929 -4547.47 4 -2050 2947 -3989.63 5 -2050 1974 -3572.70 AW is calculated by
simple division of PV by number of years, though the annuity method would have been more
exact. PV(1)=(-2050/1.104)-11500+(7705/1.104^1) -6377.72 PV(2)=(-2050/1.104)+(-
2500/1.104^2)-11500+(5239/1.104^2) -11109.6 PV(3)=(-2050/1.104)+(-2050/1.104^2)+(-
2050/1.104^3)-11500+3929/(1.104^3) -13642.4 PV(3)=(-2050/1.104)+(-2050/1.104^2)+(-
2050/1.104^3)+(-2050/1.104^4)-11500+2947/(1.104^4) -15958.5 PV(3)=(-2050/1.104)+(-
2050/1.104^2)+(-2050/1.104^3)+(-2050/1.104^4)+(-2050/1.104^5)-11500+1974/(1.104^4) -
17863.5 ESL is highest at 5th year