AP Calculus BC

Does the Sum of 1/(4n^2-1) Converge? Yes, to 1/2

The sum of 1 over 4n squared minus 1 converges to exactly 1/2. Factoring the denominator as 2n minus 1 times 2n plus 1 makes it telescope, and every piece cancels except the leading one half times 1.

∑n=1∞14n2−1\sum_{n=1}^{\infty}\frac{1}{4n^{2}-1}

Converges

sum=12\text{sum} = \frac{1}{2}

Settled by telescoping partial sums.

Factor first, and the structure appears

Written as 4n2−14n^{2}-1, the denominator hides everything. It is a difference of squares.

4n2−1=(2n−1)(2n+1)4n^{2} - 1 = (2n-1)(2n+1)

Two distinct linear factors mean partial fractions applies, and the two pieces are the same expression one odd number apart.

1(2n−1)(2n+1)=12(12n−1−12n+1)\frac{1}{(2n-1)(2n+1)} = \frac{1}{2}\left(\frac{1}{2n-1} - \frac{1}{2n+1}\right)

Only the first piece survives

In the finite sum, the −12n+1-\frac{1}{2n+1} from one term meets the +12n−1+\frac{1}{2n-1} from the next.

SN=12[(1−13)+(13−15)+⋯+(12N−1−12N+1)]=12(1−12N+1)S_{N} = \frac{1}{2}\left[\left(1 - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \cdots + \left(\frac{1}{2N-1} - \frac{1}{2N+1}\right)\right] = \frac{1}{2}\left(1 - \frac{1}{2N+1}\right)

The gap is one step through the odd numbers, so a single term is stranded at the front and a single term at the back. The back one dies in the limit.

lim⁡N→∞12(1−12N+1)=12\lim_{N \to \infty}\frac{1}{2}\left(1 - \frac{1}{2N+1}\right) = \frac{1}{2}

The mistakes students make

Two of these are algebra slips and one is a strategy slip.

  • Splitting without the 12\frac{1}{2}. Recombine and check: 12n−1−12n+1\frac{1}{2n-1} - \frac{1}{2n+1} equals 24n2−1\frac{2}{4n^{2}-1}, twice the term you started with.
  • Leaving the denominator unfactored and settling for a comparison. Comparison with 1n2\frac{1}{n^{2}} does prove convergence, but it never delivers the value 12\frac{1}{2}.
  • Reading the partial sum as 12−12N+1\frac{1}{2} - \frac{1}{2N+1}. The factor of 12\frac{1}{2} multiplies the whole bracket, so the second piece is 12(2N+1)\frac{1}{2(2N+1)}. The limit survives the slip, but every partial sum is wrong: this gives S1=16S_{1} = \frac{1}{6} when the first term is 13\frac{1}{3}.

Not sure which test a series wants?

The Convergence Test Chooser walks the decision in order: nth term first, then geometric and p-series pattern matching, then alternating structure, then the ratio test, and finally the comparison family.

Frequently asked questions

Does the sum of 1/(4n^2-1) converge?

Yes, to exactly 12\frac{1}{2}. After factoring, the series telescopes.

How do I spot a telescoping series?

Factor the denominator. If it breaks into two linear factors whose partial fractions are the same expression at shifted indices, consecutive terms cancel.

Is 1/(4n^2-1) a p-series?

No. It behaves like 14n2\frac{1}{4n^{2}} for large nn, which is enough for a verdict, but only telescoping produces the exact total.