AP Calculus BC

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

Q: How do I recognise a series with a known closed form?

Match it against the standard Maclaurin series for e^{x}, \sin x, \cos x, 1/(1-x) and \ln(1+x). A series with a factorial or a lone power of x over n is usually one of those evaluated at a specific point.

Q: Does the ratio test give the sum?

No. It gives the verdict only. Getting a sum requires recognising the series as a known expansion, telescoping it, or summing it geometrically.

The series converges, and its sum is exactly the natural logarithm of 2. The ratio test gives a limit of 1/2, and the exact value comes from the Maclaurin series for the logarithm evaluated at one half.

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

Converges

Settled by the ratio test.

The ratio test verdict

\left|\frac{a_{n+1}}{a_n}\right| = \frac{n\,2^{n}}{(n+1)\,2^{n+1}} = \frac{1}{2}\cdot\frac{n}{n+1} \longrightarrow \frac{1}{2}

The limit is $1/2 < 1$ , so the series converges absolutely. The factor $n/(n+1)$ tends to 1 and does not affect the limit, which is the usual pattern when a polynomial rides alongside an exponential.

Where the exact sum comes from

The Maclaurin series $-\ln(1-x) = \sum_{n=1}^{\infty} \frac{x^{n}}{n}$ converges for $-1 \le x < 1$ . Setting $x = 1/2$ turns the left side into $-\ln(1/2) = \ln 2$ and the right side into exactly this series.

\sum_{n=1}^{\infty}\frac{1}{n\,2^{n}} = -\ln\!\left(1 - \tfrac{1}{2}\right) = \ln 2 \approx 0.6931

It is worth noticing that $x = 1/2$ is comfortably inside the interval of convergence, so no endpoint check is needed. At $x = 1$ the same series becomes the harmonic series and diverges, which is the boundary this one stays well clear of.

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

How do I recognise a series with a known closed form?

Match it against the standard Maclaurin series for $e^{x}$ , $\sin x$ , $\cos x$ , $1/(1-x)$ and $\ln(1+x)$ . A series with a factorial or a lone power of x over n is usually one of those evaluated at a specific point.

Does the ratio test give the sum?

No. It gives the verdict only. Getting a sum requires recognising the series as a known expansion, telescoping it, or summing it geometrically.