Show that the infinite series
$u_{1} + u_{2} + u_{3} + u_{4} + . . . . . .$
is convergent or divergent according as $lim n \to \infty \sqrt[n]{u_{n}}$ is $< 1$ or $> 1$ .

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Solution

Let $u_{n}$ be the $n^{t h}$ term of the series.
Let for all $n \geq N$ ( $N =$ a natural number)
$\sqrt[n]{| u_{n} |} \leq k \leq 1$
$| u_{n} | \leq k^{n} \leq 1$
The geometric series $\sum_{n = N}^{\infty} k^{n}$ converges at $k < \frac{1}{2}$
$\sum_{n = N}^{\infty} | u_{n} |$ also converges by comparison test
Hence
If \underset{n\rightarrow \infty}{\lim}\sqrt[n]{| u_n|} \lt 1$, the series converges
And If $\sqrt[n]{| u_{n} |} > 1$ for $n$ tends to infinity, then $u_{n}$ fails to converge, hence the series is divergent.

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