Question

find the sum of series 1/(1+12+14)+2/(1+22+24)+3/(1+32+34)+.........till infinity (answer with description)?

Answer 1

Given :

The series , 1/(1+1^2+1^4)+2/(1+2^2+2^4)+3/(1+3^2+3^4)+........

To Find:

The sum of the given series till infinity.

Solution:

The n th term in the series can be written as:

• S(n) = n/(1 + n² + $n^{4}$)

• 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 + $n^{4}$) = 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 $\frac{(n^{2} + n + 1) - (n^{2} - n + 1)}{(n^{2} + n + 1)(n^{2} - n +1 )}$ = 0.5 x $\frac{1}{n^{2} - n +1 }$ - $\frac{1}{n^{2} + n +1}$

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