n-Term Test

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Calculus

Cegep/2

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189 words

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2 minutes

Test to determine whether a series diverges
Special case of limit comparison test where $b_n=1$

If $\underset{n\to \mathrm{\infty }}{lim}{a}_n\ne 0$, $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ diverges.

IMPORTANT

If $\underset{n\to \mathrm{\infty }}{lim}{a}_n=0$, $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ doesn't necessarily converge.
To prove it converges, use limit comparison test.

Proof

Assuming $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ is convergent, show $\underset{n\to \mathrm{\infty }}{lim}{a}_n=0$.

If $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ is convergent, then $\underset{N\to \mathrm{\infty }}{lim}{S}_N=S$ exists.
Since $S_N-S_{N-1}=a_N$,

$$\begin{array}{rl}\underset{N\to \mathrm{\infty }}{lim}{a}_N& =\underset{N\to \mathrm{\infty }}{lim}{S}_N-S_{N-1}\\ & =\underset{N\to \mathrm{\infty }}{lim}{S}_N-\underset{N\to \mathrm{\infty }}{lim}{S}_{N-1}\\ & =S-S\\ & =0\\ \therefore \underset{n\to \mathrm{\infty }}{lim}{a}_n& =0\end{array}$$

By contrapositive, if $\underset{n\to \mathrm{\infty }}{lim}{a}_n\ne 0$, $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ is divergent.

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Contributors

Ajitani Hifumi

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