what is sum of series 1+2^2+3^2+4^2+5^2+......+infinity
Answer must be in numerical value not formula 1/6 n (n+1)(2n+1).
Hint: As sum of infinity GP is a / 1 - r
Here the difference is in A.P. we have find numerical value of the series only.
Answers
Answer:
In this GP, if we put infinity in the place of n in the formula for calculating the sum of this GP, (1^2+2^2+3^2+....+infinity)
The answer will be infinity itself .
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Answer:
Let :
S_R = 1² + 2² + 3² + 4² + 5² + 6² + 7² + ...............
Let us take other series :
S₁ = 1² - 2² + 3² - 4² + 5² - 6² + 7² - 8² + .........
# Only applicable when number of terms are even .
Now using :
a² - b² = ( a + b ) ( a - b )
Taking two pair of even and odd term :
S₁ = ( 1 - 2 ) ( 1 + 2 ) + ( 3 - 4 ) ( 3 + 4 ) + ( 5 - 6 ) ( 5 + 6 ) + ( 7 - 8 ) ( 7 + 8 ) .......
= > S₁ = ( - 1 × 3 ) + ( - 1 × 7 ) + ( - 1 × 11 ) + ( -1 × 15 ) + ...........
Taking out - 1 as common :
= > S₁ = - 1 ( 3 + 7 + 11 + 15 + ......... )
Now rewriting 3 as 1 + 2 , 7 as 3 + 4 , 11 as 5 + 6 and so on
= > S₁ = - 1 ( 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + ......... )
From Ramanujan series for sum of infinite natural number we know :
1 + 2 + 3 + 4 + 5 + 6 + 7 + ..................... = - 1 / 12
Using this value we get :
= > S₁ = - 1 ( - 1 / 12 )
= > S₁ = 1 / 12 .
Now subtracting S₁ series from S_R i.e. S_R - S₁
S_R = 1² + 2² + 3² + 4² + 5² + 6² + 7² + 8² ...............
S₁ = 1² - 2² + 3² - 4² + 5² - 6² + 7² - 8² + ..............
- + + + +
S_R - S₁ = 8 + 32 + 72 + 128 + ..................
Take out 8 common we get :
S_R - S₁ = 8 ( 1 + 4 + 9 + 16 + .................. )
Rewrite 1 , 4 , 9 , 16 and so on as 1² , 2² , 3² 4² so on
S_R - S₁ = 8 ( 1² + 2² + 3² + 4² + ...................... )
But S_R = 1² + 2² + 3² + 4² + .....................
Replacing value by S_R
S_R - S₁ = 8 S_R
8 S_R - S_R = - S₁
7 S_R = - S₁
We have value of S₁ , putting here :
i.s. S₁ = 1 / 12
7 S_R = - 1 / 12
S_R = - 1 / ( 12 × 7 )
S_R = - 1 / 84
Therefore we get value of S_R :