AP Calculus BC

Does the Sum of (-1)^n·n/2^n Converge? Yes, to -2/9

Q: Do the terms have to decrease from the very first one?

No. The alternating series test only needs the sizes to decrease eventually. Finitely many terms at the start can do anything without affecting convergence.

Q: Why fold the sign into x rather than use the alternating test?

Because it gives the exact sum. The alternating series test only ever returns a verdict, while recognising the series as a known expansion returns the number.

The series converges absolutely, and its sum is exactly negative two ninths. Folding the sign into the base makes it a series in negative one half, and differentiating the geometric series formula gives the exact value.

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

Converges

\text{sum} = -\frac{2}{9}

Settled by the alternating series test.

Absolute convergence, then the value

Stripping the signs leaves $\sum n/2^{n}$ , which the ratio test settles with a limit of $1/2$ . So the original converges absolutely, and the alternating structure is a bonus rather than a necessity.

The terms $n/2^{n}$ rise from $1/2$ to a peak at $n = 1$ and $n = 2$ , both equal to $1/2$ , then fall away. Decreasing from the start is not required by the alternating series test either; eventually decreasing is enough.

Summing it exactly

Write the term as $n x^{n}$ with $x = -\frac{1}{2}$ , and use the identity that comes from differentiating the geometric series.

\sum_{n=1}^{\infty} n x^{n} = \frac{x}{(1-x)^{2}} \quad\Longrightarrow\quad \frac{-1/2}{(3/2)^{2}} = \frac{-1/2}{9/4} = -\frac{2}{9}

The identity holds for $|x| < 1$ , and $|-1/2| = 1/2$ is comfortably inside, so no endpoint check is needed. Note $1 - x = 1 + \frac{1}{2} = \frac{3}{2}$ ; forgetting that the minus becomes a plus is the usual slip.

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

Do the terms have to decrease from the very first one?

No. The alternating series test only needs the sizes to decrease eventually. Finitely many terms at the start can do anything without affecting convergence.

Why fold the sign into x rather than use the alternating test?

Because it gives the exact sum. The alternating series test only ever returns a verdict, while recognising the series as a known expansion returns the number.