Convergence and Divergence Math Example 1

Follow the full solution, then compare it with the other examples linked below.

Example 1

medium
Use the ratio test to determine whether ∑n=1∞n2n\displaystyle\sum_{n=1}^{\infty} \frac{n}{2^n} converges or diverges.

Solution

  1. 1
    an=n2na_n = \frac{n}{2^n}. Compute L=lim⁡n→∞∣an+1an∣L = \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|.
  2. 2
    an+1an=(n+1)/2n+1n/2n=n+1n⋅12\frac{a_{n+1}}{a_n} = \frac{(n+1)/2^{n+1}}{n/2^n} = \frac{n+1}{n} \cdot \frac{1}{2}.
  3. 3
    lim⁡n→∞n+12n=12\lim_{n\to\infty} \frac{n+1}{2n} = \frac{1}{2}.
  4. 4
    Since L=12<1L = \frac{1}{2} < 1, the series converges absolutely.

Answer

The series converges (ratio test: L=12<1L = \frac{1}{2} < 1).
The ratio test compares consecutive term sizes. L<1L < 1 means terms shrink geometrically fast enough for the sum to be finite. The series sum is actually 2 (computed via differentiation of the geometric series).

About Convergence and Divergence

A series converges if the sequence of its partial sums approaches a finite limit. A series diverges if the partial sums grow without bound or oscillate without settling.

Learn more about Convergence and Divergence →

More Convergence and Divergence Examples

Example 2 hard

Determine whether [formula] converges using the [formula]-series test.

Example 3 easy

Does [formula] converge or diverge?

Example 4 medium

Apply the ratio test to [formula].

← All Convergence and Divergence Examples Convergence and Divergence Concept