find the sum of the series,if the number of terms is infinite:
1/2.4+1/4.6/+1/6.8+1/8.10+....
sir answer 1/4
Your sum is
1/2^2 (1/1.2 + 1/2.3 + 1/3.4 + ...)
= 1/4 ∑ 1/(k(k+1))
Didn't we just do this?
To find the sum of an infinite series, we need to determine if the series converges.
In this case, we have the series 1/(2.4) + 1/(4.6) + 1/(6.8) + 1/(8.10) + ...
Let's observe the pattern of the terms in the series. We can see that each term is of the form 1/(2n)(2n+2) = 1/((2n)² + 4n).
Now, let's simplify the terms:
1/((2n)² + 4n) = 1/(4n² + 4n) = 1/(4n(n+1))
We should investigate the limit of the terms as n approaches infinity. Taking the limit of the simplified terms:
lim[n→∞] (1/(4n(n+1)))
To find this limit, we can use the concept of telescoping series.
So, let's factor out 1/4:
lim[n→∞] (1/4) * (1/n(n+1))
Now, let's break down (1/n(n+1)) into partial fractions:
1/n(n+1) = A/n + B/(n+1)
To solve for A and B, we can multiply both sides by n(n+1):
1 = A(n+1) + Bn
Expanding the equation:
1 = An + A + Bn
From here, we can equate the coefficients of like terms:
1 = (A + B)n + A
Equating coefficients, we get:
A + B = 0
A = 1
Substituting A = 1 into A + B = 0, we find that B = -1.
So, we have:
1/n(n+1) = 1/n - 1/(n+1)
Now, let's rewrite the limit using the partial fractions form:
lim[n→∞] (1/4) * (1/n - 1/(n+1))
When n approaches infinity, both 1/n and 1/(n+1) approach zero. Therefore, the limit of the terms is:
lim[n→∞] (1/4) * (0 - 0) = 0
Since the terms of the series approach zero as n goes to infinity, the series converges.
Now, let's find the sum of the series. We can write it as the difference of two series:
S = (1/4) * (1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ...)
Notice that most of the terms in the series cancel out, resulting in only the first term remaining:
S = (1/4) * 1 = 1/4
Therefore, the sum of the infinite series is 1/4.