AP Calculus BC

Does the Sum of (-1)^n/(2n+1) Converge?

Q: Is this the pi over 4 series?

It is the same terms. From n = 0 with (-1)^{n} the sum is \frac{\pi}{4}; the starting index and sign convention shift the value.

Q: How fast does it converge?

Very slowly. The error after N terms is only bounded by \frac{1}{2N+1}.

The sum of negative 1 to the n over 2n plus 1 converges CONDITIONALLY. The alternating series test applies, but the absolute values behave like the harmonic series and diverge. Written with the opposite sign convention from n equals 0, it is the Leibniz series for pi over 4.

\sum_{n=1}^{\infty}\frac{(-1)^{n}}{2n+1}

Converges

Settled by the alternating series test, and only conditionally.

The alternating series test applies

The magnitudes $\frac{1}{2n+1}$ decrease and tend to $0$ , which is all the test requires.

Absolute values give $\sum \frac{1}{2n+1}$ , which limit-comparison ties to the harmonic series, so the convergence is only conditional.

The Leibniz series for pi

Starting from $n = 0$ with the opposite sign convention gives one of the most famous series in mathematics, from the arctangent expansion at $x = 1$ .

\sum_{n=0}^{\infty}\frac{(-1)^{n}}{2n+1} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots = \frac{\pi}{4}

Beautiful and useless for computing pi

The alternating series error bound says the error after N terms is under 1/(2N+1), so five decimal places would need about fifty thousand terms. Conditional convergence is slow convergence.

The mistakes students make

Reporting the sum as $\frac{\pi}{4}$ without checking the sign convention and starting index. From $n = 1$ with $(-1)^{n}$ the value is $\frac{\pi}{4} - 1$ .
Assuming a famous series must converge quickly.

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)^n/(2n+1) converge?

Yes, conditionally, by the alternating series test.

Is this the pi over 4 series?

It is the same terms. From $n = 0$ with $(-1)^{n}$ the sum is $\frac{\pi}{4}$ ; the starting index and sign convention shift the value.

How fast does it converge?

Very slowly. The error after $N$ terms is only bounded by $\frac{1}{2N+1}$ .