SOLUTION: Find the sum of the infinite series: ∞ ∑ (3/2^i+4/5^i) i=1 So far I know that it goes: 23/10, 9

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Question 204345: Find the sum of the infinite series:

∑ (3/2^i+4/5^i)
i=1
So far I know that it goes: 23/10, 91/100, 407/1000, etc., in multiples of ten for the denominator, but I am confused as to what the top sequence is and if it's supposed to go to infinity, and then how would I find the sum?
Thank you in advance

Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!
Start with the given summation


Break up the fractions


Rewrite as . Note: for all values of 'i'


Use the identity


Break up the summation.


Re-index the summations to start them at zero. Note: this will replace each 'i' with 'i+1' since this offset occurs.


Pull out the first terms of each summation. We're subtracting them off since we're originally starting at i=1 anyway.


Raise each term to the 0th power to get 1.


Reduce


Now recall that the sum "S" for an infinite geometric series is . So this means that...


Replace the summations with the given sum formulas (see above)


Subtract


Multiply by the reciprocal


Multiply


Reduce


Combine like terms.


So


Note: if you add up the pieces of this infinite summation, you'll find that the sums will get closer and closer to 4 (you might have to find more terms).

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