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Divergence Test 📂Calculus

Divergence Test

Summary

If the series ∑n=1∞an\sum\limits_{n = 1}^{\infty} a_{n} converges, then the sequence {an}\{a_{n}\} converges to 00. ∑n=1∞an is convergent   ⟹  lim⁡n→∞an=0 \sum\limits_{n = 1}^{\infty} a_{n} \text{ is convergent } \implies \lim\limits_{n \to \infty} a_{n} = 0

Proof

Let the sum of the series be ∑n=1∞an=s\sum\limits_{n = 1}^{\infty} a_{n} = s. That is, for the partial sum sns_{n}, it is lim⁡n→∞sn=s\lim\limits_{n \to \infty} s_{n} = s. Then, since an=sn−sn−1a_{n} = s_{n} - s_{n-1},

lim⁡n→∞an=lim⁡n→∞(sn−sn−1)=lim⁡n→∞sn−lim⁡n→∞sn−1=s−s=0 \lim\limits_{n \to \infty} a_{n} = \lim\limits_{n \to \infty} (s_{n} - s_{n-1}) = \lim\limits_{n \to \infty} s_{n} - \lim\limits_{n \to \infty} s_{n-1} = s - s = 0

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Explanation

The converse is not true. In other words, if the sequence {an}\{a_{n}\} converges to 00, it does not mean that the series ∑n=1∞an\sum\limits_{n = 1}^{\infty} a_{n} converges. A well-known example is the harmonic series. The harmonic sequence {1n}\left\{ \dfrac{1}{n} \right\} converges to 00, but the harmonic series does not converge.

lim⁡n→∞1n=0 but ∑n=1∞1n=∞ \lim\limits_{n \to \infty} \dfrac{1}{n} = 0 \quad \text{ but } \quad \sum\limits_{n = 1}^{\infty} \dfrac{1}{n} = \infty

The contrapositive is the divergence test.

Divergence Test

If the sequence {an}\{a_{n}\} does not converge to 00, then the series ∑n=1∞an\sum\limits_{n = 1}^{\infty} a_{n} diverges.