AP Calculus BC

Does the Sum of n/2^n Converge? Yes, to 2

The sum of n over 2 to the n converges. The ratio test gives a limit of one half, which is less than 1. Its exact sum is 2, which comes from differentiating the geometric series rather than from any convergence test.

∑n=1∞n2n\sum_{n=1}^{\infty}\frac{n}{2^{n}}

Converges

sum=2\text{sum} = 2

Settled by the ratio test.

The ratio test settles it

an+1an=n+12n+1⋅2nn=12⋅n+1n\frac{a_{n+1}}{a_{n}} = \frac{n+1}{2^{n+1}}\cdot\frac{2^{n}}{n} = \frac{1}{2}\cdot\frac{n+1}{n}
L=lim⁡n→∞12⋅n+1n=12<1L = \lim_{n \to \infty}\frac{1}{2}\cdot\frac{n+1}{n} = \frac{1}{2} < 1

The exponential in the denominator wins against the linear numerator, so the terms shrink geometrically and the series converges.

Where the sum of 2 comes from

Differentiating the geometric series produces exactly this shape, which is how the exact value is found.

∑n=1∞nxn=x(1−x)2,∣x∣<1\sum_{n=1}^{\infty} n x^{n} = \frac{x}{(1-x)^{2}}, \qquad |x| < 1

Setting x=12x = \frac{1}{2} gives 1/2(1/2)2=2\frac{1/2}{(1/2)^{2}} = 2.

The test and the value are separate questions

The ratio test proves convergence and says nothing about the sum. Getting 2 needs a different idea entirely, and on the AP exam you would only be asked for the value if the setup pointed you at the series you should differentiate.

The mistakes students make

  • Treating it as geometric. The extra factor of nn means the ratio is not constant, only its LIMIT is 12\frac{1}{2}.
  • Reporting LL as the sum. The ratio limit is 12\frac{1}{2}; the sum is 22.

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 n/2^n converge?

Yes. The ratio test gives L=12<1L = \frac{1}{2} < 1.

What is the exact sum?

22, from the identity ∑nxn=x(1−x)2\sum n x^{n} = \frac{x}{(1-x)^{2}} evaluated at x=12x = \frac{1}{2}.

Is it geometric?

No. The factor of nn makes the ratio vary; only its limit is constant.