Question is :

If $\sum _{n=1}^{\infty} a_n$ is absolutely convergent then which of the following is not true?

  • $\sum_{m=n}^{\infty}a_m\rightarrow 0$ as $n\rightarrow \infty$
  • $\sum_{n=1}^{\infty}a_n\sin n$ is convergent.
  • $\sum_{n=1}^{\infty}e^{a_n}$ is divergent.
  • $\sum_{n=1}^{\infty}a_n^2$ is divergent.

First thing I would like to concentrate on is third option (as it is easy :P)....

absolutely convergence of $\sum _{n=1}^{\infty} a_n$ imply $a_n\rightarrow 0$ i.e., $e^{a_n}\rightarrow 1$ i.e.,$\sum_{n=1}^{\infty}e^{a_n}$ is divergent.

I guess second option is most probably true..

It is for sure absolute convergence as $|a_n\sin n|\leq |a_n|$ for all $n$.... I could not give concrete argument for convergence.

I guess fourth option is false...

absolutely convergence of $\sum _{n=1}^{\infty} a_n$ imply $a_n\rightarrow 0$ i.e., after certain stage $|a_n|<1$ i.e., $|a_n^2|<|a_n|$ So, we would have convergence of $\sum_{n=1}^{\infty}a_n^2$.

I do not understand what is actual point of first option...

Could some one confirm if this justification for second/third/fourth options is sufficient and help me to understand what first option is...

Thank you.

  • $\begingroup$ 2. Actually, you have $|a_n \sin n|\leq|a_n|$ hence $\sum a_n\sin n$ converges absolutely by comparison. 1. Note that $\sum_{m=n}^\infty a_m=\sum_{m=1}^\infty a_m- \sum_{m=1}^{n-1}a_m$. $\endgroup$ – Julien Dec 24 '13 at 5:29
  • $\begingroup$ @julien : I have already done for absolute convergence of second option.... I was looking for just convergence.... Fourth optionn I was thinking of what you have written but that was not taking me to anywhere... $\endgroup$ – user87543 Dec 24 '13 at 5:32
  • $\begingroup$ 2. You missed the absolute value. And absolute convergence implies convergence. 1. So the series $\sum_{m=1}^\infty a_m$ converges. What does this mean? $\endgroup$ – Julien Dec 24 '13 at 5:34
  • $\begingroup$ Oh yes... That was actually a typo.. :D.. absolute convergence implies convergence... Yes.. I have missed that point... I was sure that convergence need not imply absolute convergence and i was thinking absolute convergence also does not imply convergence...Face palm! $\endgroup$ – user87543 Dec 24 '13 at 5:37
  • $\begingroup$ Ah! Absolute cv does imply cv. By the Cauchy criterion. $\endgroup$ – Julien Dec 24 '13 at 5:38

$\newcommand{\ra}{\operatorname{\rightarrow}}$ I assume that the terms of your series are real numbers.

1) First notice that the absolute convergence implies that each "infinite tail" $\sum_{n=N}^{\infty} a_n$ is convergent.The condition $\sum_{m=N}^{\infty} a_m \rightarrow 0$ as $N \rightarrow \infty$ is precisely the assertion that the sequence of partial sums $\sum_{k=1}^n a_k $is a Cauchy sequence. Since $\mathbb{R}$ is complete, this is equivalent to the convergence of the series. Since absolutely convergent series are convergent, this does follow from absolute convergence.

Here the absolute convergence was used only to see that $\sum_{m=N}^{\infty} a_m$ is a real number. If we interpreted the statement as $\sup_{k \geq N} |\sum_{n=N}^{N+k} a_n| \ra 0$ as $N \ra \infty$, then we do not need absolute convergence to deduce this.

2) The inequality $|a_n \sin n| \leq |a_n|$ shows that $\sum_n |a_n \sin n| \leq \sum_n |a_n| < \infty$, so the series $\sum_n a_n \sin n$ is absolutely convergence and thus convergent.

Simple convergence of the $a_n$'s is not enough. E.g. one could take $a_n = \frac{\sin n}{n}$. It is delicate to show that this series is convergent -- one needs something like Dirichlet's Test -- but it does converge. Then $\sum_n a_n \sin n = \sum_n \frac{ \sin^2 n}{n}$, and this is divergent because e.g. $|\sin n| \geq \frac{1}{2}$ for at least $\frac{1}{3}$ of the $n$'s between $1$ and $N$.

3) What you say is correct. Notice that you used much less than absolute convergence of the series but only that $a_n \ra 0$.

4) As you say, absolute convergence implies $a_n \ra 0$, hence $a_n^2 \leq |a_n|$ for sufficiently large $n$, hence by comparison $\sum_n a_n^2$ is convergent.

Simple convergence of the $a_n$'s is not enough: take $a_n = \frac{(-1)^n}{n^{\frac{1}{2}}}$.

  • $\begingroup$ I am very happy that this has question got your view... I would say i am a big fan of you... :D keeping that aside...I am very happy with your answer and i like "Simple convergence of the $a_n$'s is not enough" part... Though i did not understand immediately how did you got that example of simple convergence is not enough, I would like to work on that...Thank you so much for sparing your time... $\endgroup$ – user87543 Dec 25 '13 at 5:31

The last one is not true since for large $n$, $|a_n| < 1$ so $a_n^2 < |a_n|$ for all such $n$.

The rest are true. Any series that is convergent must have its terms go to zero. This is true by the cauchy criterion. Choose $\epsilon > 0$. Pick $N$ so $m, n \ge N\implies |s_m - s_n| < \epsilon$, where the $s_n$ are the partial sums. Let $m = n+1$; we have $|a_{n+1}| < \epsilon$ for $n\ge N$.

  • 1
    $\begingroup$ I don't understand your intention... That part I have already proved.... $\endgroup$ – user87543 Dec 24 '13 at 5:20
  • $\begingroup$ I am familiar with what all you have written and i have already written in my question... Please help me to see how could this imply that any of the above options is true/false... $\endgroup$ – user87543 Dec 24 '13 at 5:37

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