Let $\displaystyle\sum_{n=1}^{\infty}a_n$ and $\displaystyle\lim_{n\to\infty}na_n=\infty$ and $a_1=-1$. Prove the series does not converge.

From the given that $a_1=-1$ we know that there has to be at least one more negative element because the limit is positive infinity. I don't know how to use that though.

Trying something else, after applying the condensation test, we get: $2^na_{2^n}$

Obviously: $na_n<2^na_{2^n}$ so from direct comparison the original series diverges.

It's strange that I didn't use the given $a_1=-1$. Is there another way that does use this given?

  • 1
    $\begingroup$ it is not strange, said condition is irrelevant. Formally if you want to "use" it, whatever you say, you may also precede it with "because $a_1=-1$". $\endgroup$ – Mirko Apr 11 '15 at 12:05
  • $\begingroup$ I would say $a_1$ is irrelevant, except to remind you not to say $na_n \gt k$ for all $n$ for some positive $k$ (from which you might then conclude the partial sum was greater than $kH_n$ and so diverges) $\endgroup$ – Henry Apr 11 '15 at 12:08
  • $\begingroup$ Where is the problem from? $\endgroup$ – Jonas Meyer Apr 11 '15 at 12:09
  • $\begingroup$ I found it in some file with many other problems @JonasMeyer $\endgroup$ – shinzou Apr 11 '15 at 12:14

Since $\lim na_n$ is $+\infty$, given $L>0$, there exists an index $n_0$ such that from $n \geq n_0$ we have $na_n >L$. For $L=1$ you get that from a point on we have $a_n > 1/n$. Comparing $\sum a_n$ with the harmonic series gives you the result.

The hypothesis on $a_1$ is irrelevant, since the next terms do not depend on $a_1$, and the convergence of a series is independent of the first terms.


Let $b_n=\frac{1}{n}$ and since $\lim_{n\to \infty} \frac{a_n}{b_n}=\infty$ and $\sum_n b_n $ diverges then $\sum_n a_n$ also diverges by the comparison test.


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