Study the character of the following number series: $$\sum_{n=1}^{\infty}\frac{1}{4n^2-1}$$ and in the case of convergence also determine its sum.
Solution:
To study its convergence, we can try to express each term in a simpler form. Notice that $4n^2 - 1$ can be factored as $(2n + 1)(2n - 1)$, which suggests using partial fraction decomposition(*). We can write:
$$\frac{1}{4n^2 - 1} = \frac{1}{(2n + 1)(2n - 1)} = \frac{1}{2(2n - 1)} - \frac{1}{2(2n + 1)}$$
Now, the given series becomes a telescoping series after decomposition:
$$\sum_{n=1}^{\infty}\left(\frac{1}{2(2n - 1)} - \frac{1}{2(2n + 1)}\right)$$
When we expand this series, a lot of terms will cancel out, leaving only the first and last terms:
$$\left(\frac{1}{2(2\cdot1 - 1)} - \frac{1}{2(2\cdot1 + 1)}\right) + \left(\frac{1}{2(2\cdot2 - 1)} - \frac{1}{2(2\cdot2 + 1)}\right) + \dots$$
$$= \left(\frac{1}{2(1)} - \frac{1}{2(3)}\right) + \left(\frac{1}{2(3)} - \frac{1}{2(5)}\right) + \dots + \left(\frac{1}{2(n) - 1} - \frac{1}{2(n + 1) + 1}\right)$$
As we can see, each term cancels with the next term except for the first term and the last term:
$$\left(\frac{1}{2(1)} - \frac{1}{2(3)}\right) + \left(\frac{1}{2(n) - 1} - \frac{1}{2(n + 1) + 1}\right) + \dots$$
$$= \frac{1}{2(1)} - \frac{1}{2(n + 1) + 1}$$
As $n$ approaches infinity, the second term approaches zero. Thus, the sum of the series converges to $\dfrac{1}{2}$.
(*)
$$\frac{1}{(2n + 1)(2n - 1)} = \frac{A}{2n + 1} + \frac{B}{2n - 1}$$
To find $A$ and $B$, we'll multiply both sides by the denominator $(2n + 1)(2n - 1)$ to clear the fractions:
$$1 = A(2n - 1) + B(2n + 1)$$
Expanding the right side:
$$1 = 2An - A + 2Bn + B$$
Now, we can equate the coefficients of like terms on both sides:
For $n$ terms: $$2A + 2B = 0$$
For the constant terms: $$-A + B = 1$$
We have a system of two equations:
- $2A + 2B = 0$
- $-A + B = 1$
From equation 1, we can solve for $A$: $$2A = -2B$$ $$A = -B$$
Now, substitute $A = -B$ into equation 2: $$-(-B) + B = 1$$ $$B + B = 1$$ $$2B = 1$$ $$B = \frac{1}{2}$$
Substitute $B = \frac{1}{2}$ back into $A = -B$: $$A = -\frac{1}{2}$$
So, we have found that $A = -\frac{1}{2}$ and $B = \frac{1}{2}$.
Therefore, the partial fraction decomposition of the function is:
$$\frac{1}{(2n + 1)(2n - 1)} = -\frac{1}{2(2n + 1)} + \frac{1}{2(2n- 1)}$$
Is there another way to solve the series?
