4
$\begingroup$

If we know that $\sum_{n=1}^{\infty}a_n$ converges,

what (if anything) can be known about $\sum_{n=1}^{\infty}\frac{1}{a_n}$ ?

I understand that convergence means the summation adds up to a number so this is the statement I have come up with so far:

if the number $\sum_{n=1}^{\infty}a_n$ converges to is $< 0$ or $> 0$ then $\sum_{n=1}^{\infty}\frac{1}{a_n}$ converges also. otherwise if the number $\sum_{n=1}^{\infty}a_n$ converges to $= 0$, $\sum_{n=1}^{\infty}\frac{1}{a_n}$ does not converge.

  • 10
    Given that you know that $a_n \to 0$ is necessary for convergence, you can conclude that ...2011-07-28
  • 0
    Must $an → 0$ always be true, Could $an → 1 (2,3 etc.)$ not also mean convergence?2011-07-28
  • 2
    Matt: It is indeed a necessary (but not usually sufficient) condition for the series $\sum_{n=1}^{\infty} a_n$ to converge, that we have $a_n \to 0$. For note that $a_n = s_n - s_{n-1}$, where $s_n$ is the $n$-th partial sum of the series.2011-07-28
  • 4
    To continue on @Theo"s comment: the series $\sum a_n$ converges iff the **sequence** $(A_n)$ converges to a finite limit, where $A_n=a_1+\cdots+a_n$. Now, if $(A_n)$ converges to a finite limit, you know that $A_{n+1}-A_n$ converges to...2011-07-28
  • 0
    If $\sum a_n$ converges then it must **absolutely** be true that $a_n\to 0$. How are the partial sums supposed to get trapped within arbitrarily small neighborhoods of a final limit if the increments don't eventually get smaller and smaller?2011-07-28
  • 0
    To add to Geoff's comment, a/the standard example of how the condition ${a_n}\rightarrow 0$, is necessary, but not sufficient, think of $\sum \frac{1}{n}$, in which the n-th term goes to 0, but the sum diverges.2011-07-28

2 Answers 2

1

Hint: you might think about the series for $\ln 2: \sum_{n=1}^{\infty}\frac{(-1)^n}{n}$, which converges to a non-zero value.

  • 1
    Of *all* the nonzero series you could have chosen, why did you have to choose the standard-example conditionally convergent one? :)2011-07-28
  • 0
    @anon: because standard examples come to mind easily. But I guess $2^{-n}$ would have worked as well.2011-07-28
1

Note that the OP asks "what (if anything) can be known about...". If the $a_i > 0$, by the harmonic-arithmetic mean inequality, $$(\sum_{n=1}^m a_n)/m \ge ((\sum_{n=1}^m 1/a_n)/m)^{-1}$$ or $$\sum_{n=1}^m 1/a_n \ge m^2/\sum_{n=1}^m a_n$$ We thus can get a lower bound on the sum, instead of an upper bound. If $\sum_{n=1}^{\infty} a_n$ converges, this shows that $\sum_{n=1}^m 1/a_n$ grows like $m^2$.