AP Calculus BC

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

Q: Is 1/(2^n+1) a geometric series?

No. The ratio \frac{2^{n}+1}{2^{n+1}+1} depends on n. It approaches \frac{1}{2} but is never equal to it, so the series only resembles a geometric one.

The sum of 1 over 2 to the n plus 1 converges absolutely. Each term is smaller than 1 over 2 to the n, so direct comparison with a convergent geometric series settles it. The series is not itself geometric, since its ratio is not constant.

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

Converges

Settled by the direct comparison test.

It looks geometric, and it is not

A geometric series has the same ratio between every pair of consecutive terms. Test that here.

\frac{a_{n+1}}{a_{n}} = \frac{2^{n}+1}{2^{n+1}+1}

At $n = 1$ this is $\frac{3}{5}$ , at $n = 2$ it is $\frac{5}{9}$ , and it drifts down towards $\frac{1}{2}$ without ever reaching it. The ratio changes, so the geometric sum formula does not apply and there is no first term over $1 - r$ to compute.

Direct comparison finishes it in one line

Adding $1$ to a positive denominator makes the fraction smaller, and the terms are positive to begin with.

0 < \frac{1}{2^{n}+1} < \frac{1}{2^{n}} = \left(\frac{1}{2}\right)^{n}

$\sum \left(\frac{1}{2}\right)^{n}$ is geometric with $r = \frac{1}{2}$ and $|r| < 1$ , so it converges to $1$ . Every partial sum of this series is therefore capped by that bound of $1$ , and partial sums that increase but stay capped have a limit.

Compare with geometric, do not become geometric

The comparison series is geometric. The series being tested is not. The geometric total bounds this sum from above; it is not this sum.

The mistakes students make

The trap is the resemblance to a geometric series, and it catches students in two directions.

Applying the geometric sum formula and reporting a value. With a ratio that changes from $\frac{3}{5}$ to $\frac{5}{9}$ and onwards, there is no $r$ to put in the formula.
Rewriting the term as $\frac{1}{2^{n}}$ on the grounds that the $+1$ is small. The two series are not equal, and only the inequality between them is being used.
Writing the inequality backwards, as $\frac{1}{2^{n}+1} > \frac{1}{2^{n}}$ . That is false, and even if it held it would only put the terms above a convergent series, which settles nothing. Adding $1$ to the denominator makes each term smaller, and bounding above by a convergent series is what proves convergence.

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/(2^n+1) converge?

Yes, absolutely, by direct comparison with the geometric series $\sum \left(\frac{1}{2}\right)^{n}$ .

Is 1/(2^n+1) a geometric series?

No. The ratio $\frac{2^{n}+1}{2^{n+1}+1}$ depends on $n$ . It approaches $\frac{1}{2}$ but is never equal to it, so the series only resembles a geometric one.

What is the sum of 1/(2^n+1)?

The test settles convergence but never hands you a closed form. All you get for the total is that it sits below $1$ , the geometric bound, which is as precise as this course asks you to be.