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Homesequence determining convergence of series

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# sequence determining convergence of series

Theorem. Let $a_{1}\!+\!a_{2}\!+\ldots$ be any series of real terms $a_{n}$. If the positive numbers $r_{1},\,r_{2},\,\ldots$ are such that

$\displaystyle\lim_{{n\to\infty}}\frac{a_{n}}{r_{n}}\;=\;L\;\neq\,0,$ | (1) |

then the series converges simultaneously with the series $r_{1}\!+\!r_{2}\!+\ldots$

*Proof.* In the case that the limit (1) is positive, the supposition implies that there is an integer $n_{0}$ such that

$\displaystyle 0.5L\;<\;\frac{a_{n}}{r_{n}}\;<\;1.5L\quad\textrm{for }n\geqq n% _{0}.$ | (2) |

Therefore

$0\;<\;0.5Lr_{n}\;<\;a_{n}\;<\;1.5Lr_{n}\quad\textrm{for all }n\geqq n_{0},$ |

and since the series $\sum_{{n=1}}^{\infty}0.5Lr_{n}$ and $\sum_{{n=1}}^{\infty}1.5Lr_{n}$ converge simultaneously with the series $r_{1}\!+\!r_{2}\!+\ldots$, the comparison test guarantees that the same concerns the given series $a_{1}\!+\!a_{2}\!+\ldots$

The case where (1) is negative, whence we have

$\lim_{{n\to\infty}}\frac{-a_{n}}{r_{n}}\;=\;-L>0,$ |

may be handled as above.

Note. For the case $L=0$, see the limit comparison test.

## Mathematics Subject Classification

40A05*no label found*

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