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1/1+1 2 +1 4 +1/1+2 2 +2 4 +1/1+3 2 +3 4 +......... infinite find 14s

1/1+12+14 +1/1+22+24+1/1+32+34+......... infinite find 14s

Grade:11

1 Answers

Arun
25750 Points
3 years ago
The n th term in the series can be written as:
S(n) = n/(1 + n² + )
Our aim is to convert this into a form where we can cancel adjacent terms and finally reach at a numerical value.
Therefore we should convert this into term1  -  term2 form , by using partial fractions.
S(n) = n/(1 + n² + ) = n/(
S(n) = n/ (n² + 1)² - n² =
By applying a² - b² = (a + b)(a - b)
s(n) = n/ (n² + n + 1)(n² - n + 1)
Now we can write
n = 1/2 (n² + n + 1) - (n² - n + 1)
Substituting this in S(n),
S(n) = 0.5 x  = 0.5 x  - 
Therefore,
Now adding each terms for n = 1 , 2 , 3 , ....
Sum = 0.5 ( 1 - 1/3 + 1/3 - 1/7 + 1/7 - 1/13 ......)
We can observe that the last partial fraction of nth term and first partial fraction of n+1 th term gets cancelled.
This can be extended to all n as it goes till infinity.
Therefore ,
Sum = 0.5
Therefore the sum of the series till infinity is 0.5
 

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