the n-th term test for the series $\sum_{n=1}^{\infty }{1/n}$ implies that the series converges
since $\lim_{n\to\infty} 1/n =0$
but the series actually diverges. what is the error in the procedure followed?
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Martin Sleziak
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arvind
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2Do you actually mean $\sum_{i=1}^{\infty }{1/i}$? – Bernard Nov 09 '21 at 10:27
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yes @Bernard , i edited it – arvind Nov 09 '21 at 10:28
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3What's the n-th term test? – Nov 09 '21 at 10:33
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This is only a necessary, not a sufficient condition. – Peter Nov 09 '21 at 12:54
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Does this answer your question? Converse of nth term test for divergence – Jacob Manaker Nov 09 '21 at 22:43
2 Answers
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Be careful: the theorem states
If $\sum a_n$ converges, then $\lim_{n\to \infty}{a_n}=0$
which conversely means
If $\lim_{n\to \infty}{a_n}\neq 0$ then $\sum a_n$ does not converge.
This means that $\lim_{n\to \infty}{a_n}=0$ is a necessary but not sufficient condition for a series to converge.
b00n heT
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If the series $\sum_{n=1}^{\infty} a_n$ converges, than limit of $n$-th term $a_n$ is zero as $n\to\infty$. It's because convergence means existence of $\lim_{n\to\infty} S_n$, where $S_n = \sum_{k=1}^n a_k$, and $a_n = S_{n} - S_{n-1}$, so $\lim_{n\to\infty} a_n = \lim_{n\to\infty} S_n - \lim_{n\to\infty} S_{n-1} = 0$ as latter two limits are equal. However, converse isn’t true as your example shows.
Yalikesifulei
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