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# plz help

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Find the infinite sum of
1/7 + 2/7^2 + 3/7^3 + 1/7^4 + 2/7^5 + 3/7^6 + ...

Jun 24, 2022

#1
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sumfor(n, 1, infinity, (n / (7^n))==converges to 7 / 36

Jun 24, 2022
#2
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FORGET THE ABOVE!  IT IS WRONG.

Guest Jun 24, 2022
#3
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∑[ 66 / 343] * [1 / 343]^n, n, 1, ∞]==It converges to 11 / 19551

Jun 24, 2022
#4
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Just sum it up as an infinite sequence:

First term=1/7 + 2/7^2 + 3/7^3 =66 / 343

Common Ratio =1 / 343

Sum=(66/343) / (1 - 1/343) =11 / 57 - the answer!

Jun 25, 2022
#5
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We can write the above series in a slightly different manner

(1/7) ( 1 + 1/7^3 +  1/7^6 + .......+ 1/7^(3n-3))   + (2/7^2) (1 + 1/7^3 + 1/7^6 +......) +  (3/7^3) ( 1 + 1/7^3 + 1/7^6 + ....)  =

(1 + 1/7^3  +1/7^6 +.....+ 1/7^(3n - 3) )  ( 1/7  + 2/7^2 + 3/7^3 +......+ n / 7^n )

The first series is a geometic sum with r =  1/7^3   and a first term of 1

The sum of this series =  1 / ( 1 - 1/7^3)  =  343/342

For the second series.. let the sum = S and multiply multiply each term  by r  = 1/7

So

Sr  =   1/7^2  + 2/7^3  +   ... +  n /7^n

So

S  =  1/7 + 2/7^2 + 3/7^3   +  n / 7^n

Sr =            1/7^2  +  2/7^3     + (n-1) / 7^n +  n / 7^n

S - Sr  =  1/7  + 1/7^2 + 1/7^3  +  1/7^n -  n/7^n

The term in red  approaches 0  as n approaches infinity so we can  ignore it

So

S - Sr  =  1/7 + 1/7^2 + 1/7^3 + 1/7^n

S(1 - r)  =  the sum of  a geometric  series  with r = 1/7  and a first term of 1/7

S ( 1  - 1/7)  =  (1/7) / ( 1 - 1/7)

S (6/7) =  (1/7)/ (6/7) =     1/6

S =   (1/6)(7/6) = 7 /36

So.....the sum of the  given  series =  (343/342) ( 7/36) =   2401 / 12312

Jun 25, 2022
#6
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