# Convergence of $\sum \frac{1}{a_n}$ given convergence of $\sum a_n$

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.

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Given that you know that $a_n \to 0$ is necessary for convergence, you can conclude that ... – t.b. Jul 28 '11 at 15:44
Must $an → 0$ always be true, Could $an → 1 (2,3 etc.)$ not also mean convergence? – Matt Jul 28 '11 at 15:49
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. – Geoff Robinson Jul 28 '11 at 15:52
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... – Did Jul 28 '11 at 15:54
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? – anon Jul 28 '11 at 15:55

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.
@anon: because standard examples come to mind easily. But I guess $2^{-n}$ would have worked as well. – Ross Millikan Jul 28 '11 at 16:38
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 $m^2$.